
Class _T4^^ 
Book._ 



^■4 n 



Copightl^? 



COPYRIGHT DEPOSnV 



APPLIED MECHANICS FOR ENGINEERS 



■>? 



h§?>^« 



THE MACMILLAN COMPANY 

NEW YORK • BOSTON • CHICAGO 
ATLANTA • SAN FRANCISCO 

MACMILLAN & CO., Limited 

LONDON • BOMBAY • CALCUTTA 
MELBOURNE 

THE MACMILLAN CO. OF CANADA, Ltd. 

TORONTO 



APPLIED MECHANICS 

FOR ENGINEERS 



A TEXT-BOOK FOR ENGINEERING 

STUDENTS 



BY 



E. L. HANCOCK 

ASSISTANT PROFESSOR OF APPLIED MECHANICS 
PURDUE UNIVERSITY 



Weto gorft 

THE MACMILLAN COMPANY 

1909 

All rights reserved 







^^' 



A1- 



LIBRARY of CONGRESS 
Two Cooles Received 

JAN 12 1909 

Cepyrltfttt Entry 



Copyright, 1909, 
By the MACMILLAN COMPANY. 



Set up and electrotyped. Published January, 1909. 



NorfajootJ Pregg 

J. 8. Cushing Co. — lieiwick & Smith Co. 

Norwood, Mass., U.S.A. 



a 



/<^1 



PREFACE 

In the preparation of this book the author has had in 
mind the fact that thQ student finds much difficulty in 
seeing the applications of theory to practical problems. 
For this reason each new principle developed is followed 
by a number of applications. . In many cases these are 
illustrated, and they all deal with matters that directly 
concern the engineer. It is believed that problems in 
mechanics should be practical engineering work. The 
author has endeavored to follow out this idea in writing 
the present volume. Accordingly, the title "Applied 
Mechanics for Engineers " has been given to the book. 

The book is intended as a text-book for engineering 
students of the Junior year. The subject-matter is such 
as is usually covered by the work of one semester. In 
some chapters more material is presented than can be used 
in this time. With this idea in mind, the articles in 
these chapters have been arranged so that those coming 
last may be omitted without affecting the continuity of 
the work. The book contains more problems than can 
usually be given in any one semester. 

While it is difficult to present new material in the 
matter of principles, much that is new has been intro- 
duced in the applications of these principles. The sub- 
ject of Couples is treated by representing the couples 
by means of vectors. The author claims that the chap- 
ters on Moment of Inertia, Center of Gravity, Work 
and Energy, Friction and Impact are more complete 
in theory and applications than those of any other 



vi PREFACE 

American text-book on the same subject. These are 
matters upon which the engineer frequently needs infor- 
mation ; frequent reference is, therefore, given to origi- 
nal sources of information. It is hoped that these chapters 
will be especially helpful to engineers as well as to students 
in college, and that they will receive much benefit as 
a result of looking up the references cited. In general, 
the answers to the problems have been omitted for the 
reason that students who are prepared to use this book 
should be taught to check their results and work inde- 
pendently of any printed answer. 

The author wishes to acknowledge the helpful sugges- 
tions obtained from the many standard works on me- 
chanics. An attempt has been made to give the specific 
reference to the original for material taken from engi- 
neering works or periodical literature. He wishes, more- 
over, to express his thanks to Dean C. H. Benjamin and 
Professor L. V. Ludy for their careful reading of the 
manuscript, to Professor W. K. Hatt for many of the 
problems used, and to Dean W. F. M. Goss, whose con- 
tinued interest and advice have been a constant source of 
inspiration. It is hoped that the work may be an inspira- 
tion to students of engineering. 

E. L. HANCOCK. 

Purdue University, 
November, 1908. 



TABLE OF CONTENTS 

CHAPTER I 
ARTICLES 1-15 

PAOS 

Definitions 1 

Introduction — Force — Unit of Force — Unit Weight 
— Rigid Body — Inertia — ]\Iass — Displacement — Repre- 
sentation of Forces — Concurrent Forces — Resultant of 
Two Concurrent Forces — Resolution of Force — Force 
Triangle — Force Polygon — Transmissibility of Forces. 

CHAPTER II 

ARTICLES 16-19 

Concurrent Forces 9 

Concurrent Forces in a Plane — Concurrent Forces in 
Space — Moment of a Force — Varignon's Theorem of Mo- 
ments. 

CHAPTER III 

ARTICLES 20-21 

Parallel Forces 20 

Parallel Forces in a Plane — Parallel Forces in Space. 

CHAPTER IV 

ARTICLES 22-29 

Center of Gravity 27 

Definition of Center of Gravity — Center of Gravity de- 
termined by Symmetry — Center of Gravity determined by 
Aid of the Calcuhis — Center of Gravity of Locomotive Coun- 
terbalance — Simpson's Rule — Application of Simpson's 
Rule — Durand's Rule — Theorems of Pappus and Guldinus. 

vii 



Viil CONTENTS 

CHAPTER V 
ARTICLES 30-34 

PAGE 

Couples 50 

Couples Defined — Representation of Couples — Couples 
in One Plane — Couples in Parallel Planes — Couples in In- 
tersecting Planes. 

CHAPTER VI 

ARTICLES 35-36 

Non-concurrent Forces 56 

!N'on-concurrent Forces in a Plane — Non-concurrent 
Forces in Space. 

CHAPTER VII 

ARTICLES 37-65 

Moment of Inertia 69 

Definition of Moment of Inertia — Meaning of Term — 
Units of Moment of Inertia — Representation of Moment 
of Inertia — Moment of Inertia, Parallel Axes — Inclined 
Axes — Product of Inertia — Axes of Greatest and Least 
Moment of Inertia — Polar Moment of Inertia — Moment 
of Inertia of a Rectangle — Triangle — Circular Area — 
Elliptical Area — Angle Section — Moment of Inertia by 
Graphical Method — jMoment of Inertia by Use of Simpson's 
Rule — Least Moment of Inertia of Area — The Ellipse of 
Inertia — IMoment of Inertia of Thin Plates — Right Prism 
— With Respect to Geometrical Axes — Of Solid of Revo- 
lution — Of Right Circular Cone — IMoment of Inertia of 
]\Iass — Moment of Inertia of Xon-honiogeneous Bodies — 
Of Mass, Inclined Axis — Principal Axes — El]i]3soid of 
Inertia — Moment of Inertia of Locomotive Drive Wheel. 

CHAPTER VIII 

ARTICLES 66-70 
Flexible Cords Ill 

Introduction — Cords and Pulleys — Cord with Uniform 
Load Horizontally — Equilibrium of Cord due to its Own 
Weight — Representation by Means of Hyperbolic Function. 



CONTENTS ix 

CHAPTER IX 
ARTICLES 71-84 

PAGE 

Rectilinear Motion 123 

Velocity — Acceleration — Constant Acceleration — Freely 
Falling Bodies — Bodies Projected vertically Upward — 
Newton's Laws of Motion — Motion on an Inclined Plane 

— Variable Acceleration — Harmonic Motion — Motion 
with Repulsive Force Acting — Resistance varies as Dis- 
tance — Attractive Force varies as Square of Distance — 
Motion of Body falling through Atmosphere — Relative 
Velocity. 

CHAPTER X 

ARTICLES 85-94 

Curvilinear Motion 142 

Representation of Velocity and Acceleration — Tangen- 
tial and Normal Accelerations — Simple Circular Pendulum 

— Cycloidal Pendulum — JMotion of Projectile in Vacuo — 
Body projected up an Inclined Plane — Motion of Projec- 
tile in Resisting Medium — Path of Projectile, Small Angle 
of Elevation — Motion in Twisted Curve. 

CHAPTER XI 

ARTICLES 95-100 

Rotary Motion 169 

Angular Velocity — Angular Acceleration — Constant 
Angular Acceleration — Variable Acceleration — Combined 
Rotation and Translation — Rotation in General. 



CHAPTER XII 

ARTICLES 101-128 

Dynamics of Machinery 177 

Statement of D'Alembert's Principle — Simple Transla- 
tion of a Rigid Body — Simple Rotation of Rigid Body — 



X CONTENTS 

PAGE 

Reactions of Supports ; Rotating Body — Rotation of a 
Sphere — Center of Percussion — Compound Pendulum — 
Experimental Determination of Moment of Inertia — Deter- 
mination of g — The Torsion Balance — Constant Angular 
Velocity — Rigid Body Free to Rotate — Rotation of Sym- 
metrical Bodies — Rotation of Locomotive Drive Wheel — 
Rotation about an Axis not a Gravity Axis — Rotation of 
Fly Wheel of Steam Engine — Rotation and Translation — 
Side Rod of Locomotive — The Connecting Rod — Body 
rotating about an Axis, One Point Fixed — Gyroscope — 
The Spinning Top — Motion of Earth — Plane of Rotation 

— Gyroscopic Action Explained — Precessional Moment, 
Special Case — General Case — Car on Single Rail. 

CHAPTER XIII 

ARTICLES 129-144 

Work and Energy 229 

Definitions — Units of Work — Graphical Representation 
of Work — Power — Energy — Conservation of Energy — 
Energy of Body moving in Straight Line — Work under 
Action of Variable Force — Pile Driver — Steam Hammer 

— Energy of Rotation — Brake Shoe Testing Machine — 
Work of Combined Rotation and Translation — Kinetic 
Energy of Rolling Bodies — Work-Energy Relation for Any 
Motion — Work done when Motion is Uniform. 



CHAPTER XIV 

ARTICLES 145-168 

Friction 261 

Friction — Coefficient of Friction — Laws of Friction — 
— Friction of Lubricated Surfaces — ^lethodof Testing Lu- 
bricants — Rolling Friction — Friction W^heels — Resistance 
of Ordinary Roads — Roller Bearings — Ball Bearings — 
Friction Gears — Friction of Belts — Transmission Dyna- 
mometer — Creeping of Belts — Coefficient of Friction of 
Belts — Centrifugal Tension of Belts — Stiffness of Belts 



CONTENTS XI 

PAGE 

and Ropes — Friction of Worn Bearing — Friction of Pivots : 
Flat Pivot, Collar Bearing, Conical Pivot, Spherical Pivot 
— Absorption Dynamometer — Friction Brake — l^rony Fric- 
tion Brake — Friction of Brake Shoes — Train Kesistance. 



CHAPTER XV 

ARTJCLFS 109-178 

Impact 315 

Definitions — Direct Central Impact, Inelastic; Elastic 
— Elasticity of Materials — Impact of Imperfectly Elastic 
Bodies — Impact Tension and Impact Compression — Di- 
rect Eccentric Impact — Center of Percussion — Oblique 
Impact of Body against Smooth Plane — Impact of Rotating 
Bodies. 

Appendix I. Hyperbolic Functions, Tables. 

Appendix II. Logarithms of Xumbers. 

Appendix III. Trigonometric Functions, Tables. 

Appendix IV. Squares, Cubes, Square Roots, etc., of Numbers. 

Appendix V. Conversion Tables. 

Index 383 



APPLIED MECHANICS FOR ENGINEERS 



CHAPTER I 

DEFINITIONS 

1. Introduction. — The study of the subject of mechanics 
of engineering involves a study of matter^ space^ and time. 
The subject as presented in tliis book consists of two parts ; 
viz., statics, including the study of bodies under the action 
of systems of forces that are in equilibrium (balanced), 
and dynamics, including a study of the motion of bodies. 

2. Force. — A body acted upon by the attraction or 
repulsion of another body is said to be subjected to an 
attractive or repulsive force, as the case may be. Forces 
are usually defined by the effects produced by them, as 
for example, we say, a force is something that produces 
motion or tends to produce motion, or changes or tends 
to change motion, or that changes the size or shape of a 
body. The study of relations between forces and the 
motions produced by them is usually designated as the 
study of Statics and Dynamics, Forces always occur in 
pairs ; for example, a book held in the outstretched hand 
exerts a downward pressure on the hand, and the hand 
exerts an equal upward pressure on the book. 

3. Unit of Force. — The unit of force used by engineers 
in this country and England is the pound avoirdupois. It 



2 APPLIED MECHANICS FOE ENGINEEBS 

is sometimes, however, necessary to use the absolute unit 
of force. This ijaay be defined as follows : The absolute 
unit of force is .that force which acting on a unit mass 
during unit time will produce in the mass, unit velocity.] 
This absolute unit of force is called a poundal. In France, 
Germany, and other countries where the centimeter-gram- 
second system is used, the engineer's unit of force is the 
kilogram. The absolute unit of force, in such countries, 
is the force which acting upon a mass of one gram weight 
(at Paris) w^ill produce a velocity of one centimeter per 
second, in a second. Such a unit is called a dyne. 

4. Unit Weight. — The weiglit of a cubic foot of a sub- 
stance will be called the unit weight of the substance and 
will be represented by 7. Below is given a table of such 
weights taken at the sea level. It will be seen that the 
unit weight of a substance divided by the unit weight 
of pure water gives its specific gravity. (See Table I on 
opposite page.) 

5. Rigid Body. — In studying the state of motion or 
rest of a body due to the application of forces acting upon 
it, it is not necessary to consider the deformation of the 
body itself, due to the forces. When so considered it is 
customary to say that the body is a rigid body. Unless 
otherwise stated bodies will be considered as rigid bodies 
in this book. 

6. Inertia. — The property of a body that causes it to 
continue in motion, if in motion, or remain at rest, if at rest, 
unless acted upon by some other force, is called inertia. 
This is Newton's First Law of Motion. (See Art. 76.) 



DEFINITIONS 



TABLE I 
Unit Weights and Specific Gravity of Some Materials 

(Kent's *' Engineer's Pocket Book ") 



Material 


Specific 


Gravity 


Unit Weight 


Brass 


8.2 to 8.G 


511 to 536 


Brick 








Soft 


1.6 




100 


Common 


1.79 




112 


Hard 


2.0 




125 


Pressed 


2.16 




135 


Fire 


2.24- 


-2.4 


140-150 


Brickwork — mortar 


1.6 




100 


Brickwork — cement 


1.79 




112 


Concrete 


1.92- 


-2.24 


120-140 


Copper 


8.85 




552 


Earth — loose 


1.15- 


-1.28 


72-80 


Earth — rammed 


1.44- 


-1.76 


90-110 


Granite 


2.56- 


■2.72 


160-170 


Gum 


.92 




57 


Hickory 


.77 




48 


Iron — cast 


7.21 




450 


Iron — wrought 


7.7 




480 


Lead 


11.38 




709.7 


Limestone 


2.72- 


3.2 


170-200 


Masonry — dressed 


2.24- 


2.88 


140-180 


Nickel 


8.8 




548.7 


Pine— white 


.45 




28 


Pine— yellow 


.61 




38 


Poplar 


.48 




30 


Sandstone 


2.24-i 


2A 


140-150 


Steel 


7.85 




490 


White Oak 


.77 




48 



7. Mass. —The mass of a body is the quantity of 
matter it contains. Mass differs from weight, in that 
the weight varies with the position on the surface of the 



4 APPLIED MECHANICS FOB ENGINEERS 

earth and with the height above tlie surface, while the 
mass remains the same. The engineer's definition of 
mass, viz. that it is equal to the weight divided by the 
acceleration of gravity (see Art. 76), may be expressed 

^= — • I^oth a and g vary for different localities, but 
the quotient is constant ; that is, the quantity of matter in 
a body is independent of its position with reference to the 
earth. The weight of a body may be determined by 
means of the spring balance. Such a balance is the only 
true measure of weight, since the equal-armed balance 
gives the same weight regardless of distance from the 
center of the earth. The equal-armed balance really 
measures mass. 

8. Displacement. — By the displacement of a body is 
meant its change from one position to another. A dis- 
placement involves a movement in a definite direction. 
It may be represented by an arrow, the length of the 
arrow representing the distance moved and the direction 
of the arrow the direction of the motion. Thus, if a man 
walks due east one mile and then due north one mile, we 
might represent his displacement from the original posi- 
tion by an arrow drawn northeast of a length equal to V2 
miles. Or, in Fig. 1, if P^ represents a displacement of 
a body in the direction indicated and Pj a subsequent dis- 
placement in the direction of Pj, then B represents a dis- 
placement equivalent to P^ and F^. It is seen that B 
may be determined by constructing a parallelogram on Pj 
and P2 as sides and drawing the diagonal. Quantities 
that may be represented by arrows are known as vector 
quantities, and tlie arrows themselves as vectors. 



DEFINITIONS 5 

9. Representation of Forces. — Forces have a certain 
magnitude, act in a certain direction, and have a definite 
point of application. If a man, for example, attaches a 
rope to a log and pulls on the rope, his pull may be meas- 
ured in pounds ; it acts along the rope, and it has a point 
of application which is the same as the point of attach- 
ment of the rope to the log. It has been found convenient, 
for the purpose of analysis, to represent forces by arrows 
(vectors of Art. 8), the length of the arrow representing 
the magnitude of the force and the direction of the arrow 
giving the direction in which it acts. Thus, a 10-pound 
force, acting in a direction 30° with the horizontal, is 
represented by an arrow drawn in the same direction and 
having its point of application in the body and having 
a length representing 10 lb. (In this case, if 2 lb. 
represents 1 in., the length of the arrow is 5 in.) The 
line along which a force acts will be referred to as its 
line of action, 

10. Concurrent Forces. — When two or more forces act 
upon the same point of a body, their lines of action are 
concurrent^ and the forces are known as concurrent forces, 

11. Resultant of Two Concurrent Forces. — If two forces 
having the same point of application act on a body, there 
is some single force that might be applied at the same 
point to produce the same effect. This single force is 
called the resultant of the two forces, and is found as 
follows : construct upon the arrows representing the 
forces a parallelogram and draw the diagonal from the 
point of application. This diagonal represents the re- 
sultant of the two forces in macfnitude and direction 



6 APPLIED MECHANICS FOR ENGINEEES 





Fig. 1 

(Art. 8). Thus, if P^ and P^ (Fig. 1) are the forces, 
then R is the resultant. 

Algebraically ^ = ^ r^^ + r,^ -^2I>,r,cos AOB . 

12. Resolution of Force. — We have just seen how two 
concurrent forces may be replaced by a single force called 
their resultant. In a similar way a single force may be 
resolved into two forces. These forces are the sides of a 
parallelogram of which the single force is a diagonal. It 
is clear, then, that there are an infinite number of compo- 
nents into which a single resultant may be resolved. It 
is necessary, therefore, in speaking of the components of 
a force, to state specifically which are intended. It will 
be seen in problems that follow that the components most 
often used are at right angles to each other, and usually 
the vertical and horizontal components. In such a case 
the components are the projections of the force on the verti- 
cal and horizontal lines. 

13. Force Triangle. — It follows directly from the par- 
allelogram law of forces (Art. 11) that if we draw from 
any point a line parallel to and representing one of two 
concurrent forces, P^ say, and from the extremity of this 
line another line parallel to P^ and of the same length, then 
the remaining side of the triangle will be represented by H. 
This triangle is called the force triangle. In general, the 
resultant of two concurrent forces may be found by drawing 



JDEFIXinONS 7 

lines parallel to the forces as above. The line necessary 
to complete the triangle is tlie resultant, and its arrow is 
always away from the point of application. The equal and 
opposite of this resultant would be a single force that 
would hold the two concurrent forces in equilibrium. 




14. Force Polygon. — If more than two forces are con- 
current, we may find their resultant by proceeding in a way 
similar to that outlined above. Thus, let the forces be P-^, 
P^, P3, P4, etc. (Fig. 2), all passing through a point ; from 
any point draw 
a line equal and 
parallel to P-^, 
from the ex- 
tremity of the 
line draw an- 
other equal and 
parallel to P^^ from the extremity of this last line draw 
another equal and parallel to P3, and proceed in the same 
way for the other forces. The figure produced will be a 
polygon whose sides are equal and parallel to the forces. 
The resultant will be given in magnitude, direction, and 
point of application by the line necessary to close the 
polygon. The arrow, representing the direction of the 
resultant, will always be away from the point of applica- 
tion. (See Fig. 2.) If the polygon be closed, the sj^stem 
of forces will be in equilibrium. The single force neces- 
sary to produce equilibrium will, in any case, be equal and 
opposite to P. The student should construct force poly- 
gons by taking the forces in different orders and checking 
the resultant in each case. 



8 APPLIED MECHANICS FOR ENGINEERS 

By drawing the lines OA^ OB^ OQ^ etc., it is easy to 
see that OA represents the resultant of P^ and P^, that 
OB represents the resultant of OA and Pg, and so of Pj, Pg, 
and Pg, etc. That is, it is easy to' see that the force polj^- 
gon follows directly from the force triangle. By means 
of the force polygon it is easy to find graphically the 
resultant of any number of concurrent forces in a plane. 
The work, however, must be done accurately. 

The student should show that the force polygon may be 
used for finding the resultant of concurring forces in space, 
by considering two forces at a time. The force polygon 
in this case is called a twisted polygon. 

15. Transmissibility of Forces. — It is a matter of expe- 
rience that the point of application of a force may be 
changed to any point along its line of action without 
changing the effect of the force upon the rigid body. 
This, of course, is on the assumption that all the force is 
transmitted to the body. The law may be stated as fol- 
lows : The point of application of a force may be transferred 
anywhere along its line of action without changing its effect 
upon the body upon which it acts. 



CHAPTER II 



CONCURRENT FORCES 

16. Concurrent Forces in a Plane. — It will often be 
convenient to consider forces as acting on a material 
point; this is equivalent to considering the body without 
weight and 
simply a point. 
If a material 
point (0) (Fig. 
3) be acted up- 
on by a num- 
ber of forces in 
a plane, P^, P^^ 
-T^g, x"^^, etc., 
each one mak- 
ing angles a^, 

«2^ "3' ^4' ^^^-^ 
respectively, 

with the posi- 
tive 2:-axis, it 
is desirable to find the resultant of all of them in magni- 
tude and direction; that is, the single ideal force that 
could produce the same effect as the system of forces. 

Each force P may be resolved into components along 
the a> and y-axes, giving P cos a along the rc-axis, and P 

9 




10 



APPLIED MECHANICS FOB ENGINEERS 



sin a along the ?/-axis. The sum of these components 
along the a;-axis may be expressed, 

2a; = P^ cos a^ + P^ cos a^ + Pg cos a^ + etc., 

the proper algebraic sign being given cos a in each case. 
In a similar way the sum of the components along the 
y-axis may be written, 

2y = -Pi sin a^ + P^ sin a^ + P^ sin ^3 + etc. 

These forces, ^x and 2y, may now replace the original 
system as shown in Fig. 4. And these may be combined 

into a single re- 
sultant which 
is the diagonal 
of the rectangle 
of which the 
two forces are 
sides (Art. 11). 
This gives the 
resultant in 
magnitude and 
direction, and 
this resultant 
force is the sin- 
gle force which 
if allowed to 
act upon the 
material point would produce the same effect as the 
system of forces. It should be remembered in all that 
follows that this resultant force has no real existence ; it 



sr 







sx 



->-— z 



Fig. 4 



CONCUmiENT FORCES 



11 



is used to simplify the solution of problems. Analytically 
the resultant may be expressed, 



i2 = V(2x)2+(Ey)2, 



and its direction a as such an angle that tan a = 



-^y. 



^x 



(See 



Fig. 5.) If the material point be at rest or moving 
uniformly, this resultant force must be equal to zero; 

This means that (22:)^+ (2z/)2 = 0, that is, that the sum 

of two squares y 

must be zero ; 

but this can 

happen only 

when each one, 

separately, is 2y 

zero (since 

neither can be *^' 

negative being 

squared). We 

therefore have 

as the necessary 

and sufficient 

conditions for 

the equilibrium 

of a material point, acted upon by a system of concurring 

forces in a plane, 

iJ = or 2a5 = 0, and 2?/ = 0. 

When It is not zero, the system of forces causes accel- 
erated motion in the direction of R\ when li=0, the 




12 



APPLIED MECHANICS FOR ENGINEERS 




material point remains at rest, if at rest, or continues in 
motion with uniform velocity, if in motion. In this case 

the system of forces 
is said to be balanced. 
As an illustration 
of the foregoing, 
consider the case of 
a body of weight G- 
situated on an in- 
clined plane, making 
an angle 6 with the 
horizontal. (See 

Fig. 6.) There is 
a certain force P 
making an angle cf) 
with the plane, 
whose component along the plane acts upwards, and also 
a force of friction F upwards. The other forces acting 
on the body are Gr, the force of gravity acting vertically, 
and iV, the normal pressure of the plane. Taking the 
a:-axis along the plane positive upward and the ^/-axis 
perpendicular to it positive upward, we get, 

'2x=Pcos(b-^F- asm 6, 
and 2^/ = N+ P sin (f>— Gr cos 0. 

For equilibrium 

Pcos<^ + i"- (? sin (9=0, 

N+Psmcf)- G^cos(9 = 0. 

Therefore, N= G cos — P sin <^, 

asm0-F 



Fig. 6 



P = 



COS (f> 



CONCURRENT FORCES 



13 




This last equation gives the magnitude of P required to 
preserve equilibrium, supposing that the force of friction, 
^, and ^ are known. 

Problem 1. An angle iron whose 

weight is 20 lb. and angle a right angle, 

rests upon a circular shaft, radius 2 in. 

Find the normal pressure at .4 and B 

(Fig. 7). 

Problem 2. (jiven three concurring forces, 100 lb., 50 lb., and 

200 lb., whose directions referred to the a;-axis are 0°, QO"", 180^ 

respectively ; find the 
resultant in magnitude 
and direction. 

Problem 3. A body 
(Fig. 8) whose weight 
is G is drawn up the 
inclined plane with uni- 
form velocity due to the 
action of the forces I^ 
and P', Find the force 
of friction and the 
IGO lb. P acts 




Fig. 8 



normal pressure, if P = 100 lb., P' = 100 lb., G 
parallel to the plane and P' acts horizontally. 

Problem 4. A w^heel is about to roll over an obstruction. The 
diameter of the wheel (Fig. 0) is -V and its weight 800 lb. Find 
the force P necessary to start the w^heel over the obstruction. 




14 



APPLIED MECHANICS FOR ENGINEERS 




Problem 5. A weight of 10 tons is supported as shown in 
Fig. 10. Find the force acting in the tie A and the member B, 

17. Concurrent Forces in Space. — If the material point 
(0) be acted upon by a system of concurrent forces not 

in a plane P^^ 

-^2' 3' 4' etc., 
whose direc- 
tion angles are 

etc., respec- 
tively, one of 
which is shown 
in Fig. 11, the 
resultant force 
may be found in magnitude and direction by an analysis 
similar to that used in the preceding case. The sum of 
the components of all the forces along the a;-axis is 
2 a: = Pj cos a^ + P^ cos a^ + P^ cos a^ + P^ cos a^ + etc. 

Similarly, the sum of the components along the ^/-axis, 
2y = Pj cos ySj + P^ cos ySg + -P3 cos ySg + P^ cos ^^ + etc., 




Fig. 11 



CONCURRENT FORCES 



15 



and the sum of the components along the 2-axis, 

S 2 = Pj cos 7j + Pg C^S 72 + ^3 ^^^ 73 + ^4 ^^^ 74 + ^^^• 

The original system of forces may now be replaced 
by a system of three rectangular forces 2a:, 2y, and 
tz (Fig. 12). ^ 

Finally, this 
system may be 
replaced by a 
resultant which 
is the diagonal 
of a paral- 
lelopiped con- 
structed with 
^x^ 2y, and ^z 

as edges. In P^'" Fig. 12 

magnitude this resultant may be expressed 



•sz 



zx 



R = V(2;x)2 + (2^)2 + {^z)\ (See Fig. 13) 

and its direction given by the angles a, yS, and 7. These 
angles are given by the equations 

^x _ 2y S2 

cos a = -~^ COS /3 = ~^ cos 7 = -^. 
Jit ^ -/^ 

For equilibrium jR must be ; that is, 

(22:)2 + (2y)2+ (2^)2=0, 

^x = 0, 2?/ = 0, ^z = 0. 

This gives three equations of condition from which three 
unknown quantities may be determined. In the preced- 
ing case of Art. 16 there were only two equations of 
condition ^x= and Sy = ; consequently, only two un- 
known quantities could be determined. 



and therefore. 



16 



APPLIED MECHANICS FOR ENGINEEERS 




Fig. 13 

Problem 6. Three men (Fig. 14) are each pulling with a force P 
at the points a, b, and c, respectively. What weight Q can they raise 

with uniform motion if each man pulls 100 lb.? 
Each force makes an angle of 60° with the 
horizontal. 

Problem 7. Three concurring forces act 
upon a rigid body. Find the resultant in mag- 
nitude and direction. The forces are defined 
as follows : 




Pi = 75 lb. 



a. 



Pg = 80 lb 



^3 = 147^2'; 73 = ? 



Fig. 14 



63° 27' ; )8i = 48° 36' ; y^ = 

a, = 153° 44' ; 

P. = 95 lb. ; a. = 76° 14' 

Hint, yj, yo, and yg may be found from 
either of the following relations : 

cos (a + fS) cos («-/?)+ cos^y = 0, 

C0S2« + COs2/5 + COS^y = 1. 



Problem 8. Each leg of a pair of shears (Fig. 15) is 50 ft. long. 
They are spread 20 ft. at the foot. The back stay is 75 ft. long. 
Find the forces acting on each member when lifting a load of 20 tons 
at a distance of 20 ft. from the foot of the shear legs, neglecting the 
weight of structure. 



CONCURRENT FORCES 



17 




20 TONS 



Fig. 15 



18. Moment of a Force. — The moment of a force with 
respect to any point in its plane may he defined as the prod- 
uct of the force and a 
perpendicular let fall 
from the point on the 
line of action of the 
force. Let P (Fig. 16) 
be the force and the 
point and a the perpen- 
dicular distance of the 
force from the point ; 
then Pa is the moment 
of the force with respect to the point 0. This moment 
is measured in terms of the units of both force and 
lengthy viz. foot-pounds or inch-pounds, and is read foot- 
pounds moment or inch-pounds moment to distinguish 
it from foot-pounds work or inch-pounds work. 

For convenience the algebraic sign of the moment is 




Fig. 16 



18 



APPLIED MECHANICS FOR ENGINEERS 



said to be positive when the moment tends to turn the 
body in a direction counter-clockwise^ and negative when it 
tends to turn the body in the clockivise direction. 

The moment may be represented geometrically as fol- 
lows : let EF represent the magnitude of P, drawn to 
the desired scale, and draw EO and FO, The area of the 
triangle OEF ^ {EFa, or EFa=2A0EF^ that is, the 
moment of the force with respect to a point is geometrically 
represented hy twice the area of the triangle^ whose lase is 
the line representing the magnitude of the force and ivhose 
vertex is the given point. 



19. Varignon's Theorem of Moments. — The moment of 

the resultant of two concurring forces with respect to any 

point in their plane is equal to the algebraic sum of the 

moments of the two forces with respect to the same-point. 

The given forces P and P^ may be represented by OP 

and OPj and 
their resultant 
by OR. Take 
as the point 
with respect to 
which moments 
are to be taken, 
and construct 
the dotted lines 
as shown in Fig. 
17. The rao- 
FiG. n' jnent of OP 

with respect to is twice the area of the triangle OOP 
(Art. 18), = 2 area of the triangle OOP, since the tri- 




CONCURRENT FORCES 19 

angles have the same base and the same altitude. That 
is, the moment of CP with respect to is the same as the 
moment of CB with respect to = CBa^ where a is the 
perpendicular let fall from on OR, In a similar way, it 
is seen that the moment of CP^ with respect to is equal 
to the moment of CA with respect to 0; that is, to OA • a. 
Therefore, the sum of the moments of P and P^ with 
respect to equals (OB + OA) • a. But (OB + OA)a = 
(0A + AB) 'a==R ' A, since OB = AR (equal triangles 
OPB and AP^R), When the point is taken between the 
P and Pp the moment of the resultant equals the differ- 
ence of the moments of P and P^ Let the student show 
that this is true. 

Cor. 1. If there are any number of concurring forces 
in a plane, it may be shown that Varignon's theorem holds 
by considering the resultant of two of them with the 
third, and so on. The more general theorem may then be 
stated as follows : The moment of the resultant of any num- 
ber of concurring forces in a plane with respect to any point 
in that plane is equal to the algebraic sum of the moments of 
the forces with respect to the same point. 

Cor. 2. If the point be taken in the line of action of 
iJ, then a = 0^ and therefore the sum of the positive 
moments equals the sum of the negative moments. 

The moment of a force with respect to a line at right 
angles to the line of action of the force is the product of the 
force and the shortest distance between the two lines. 

The moment of a force with respect to a line not at right 
angles to the line of action of the force is the same as the 
moment of the component of the force in a plane perpendic- 
ular to the line. 



CHAPTER III 



PARALLEL FORCES 



20. The Resultant of Two Parallel Forces. — In consid- 
ering two parallel forces in a plane three cases arise : 
(a) when the forces are in the same direction ; (5) when 
they are unequal and in opposite directions; (^) when 
thej- are equal and in opposite directions, but having dif- 
ferent lines of action. 




Fig. 18 



Case (a). When the forces are in same direction. The 
two forces are P and P^, and the distance between their 
lines of action is a. For the sake of analysis put in two 

20 



PARALLEL FORCES 21 

equal and opposite forces T as shown in Fig. 18. These 
forces will have no effect as far as the state of motion of 
the body is concerned. The resultant of T and P is iJ, 
and that of T and P^ is By Transfer It and R^ to the 
point of intersection of their lines of action A. Here 
resolve them into components parallel to their original 
components ; the two forces 2^ nullify each other, and there 
are left the two forces P and P^ acting along the same line 
AU. The resultant of P and P^ then, is equal to P + P^ 
and acts in the same direction as the forces. 

To determine the position of P with reference to the 
forces we have from similar triangles 

T 
P' 

irom which — i = , or a; = — . 

P X R 

That 18^ the resultant of two parallel forces in the same direc- 
tion divides the distance between them in the inverse ratio 
of the forces. 

Cor. For any point in the same plane, it is easy to 
show that the moment of the resultant is equal to the 
algebraic sum of the moments of the two forces with 
respect to this same point. Draw a line through paral- 
lel to T and let m be the distance from to the line of 
action of P^ If now Pm be added to both sides of the 

equation 

Px = Pa^ 

we shall have 

Ii(7n + x) = P{a + m) + P-^m^ 

and this is the relation we were to find. 



X 


T 
Pi 


and 


a — X 


AF 


AF 


Py- 


a — 


•^ n 


V Xz=- 



22 APPLIED MECHANICS FOR ENGINEERS 

When (9 is a point on the line of action of B, the 
moment of iJ = 0, and we have the moment of P equal to 
the moment of P^. This is often a convenient relation to 
use in the solution of problems. Following out the above 
reasoning, let the student show that the moment of the 
resultant of any number of parallel forces in a plane with 
respect to a point in the plane is equal to the algebraic sum 
of the moments of the forces with respect to that same point. 

Case (J). When the forces are unequal and opposite in 
direction. In this case the analysis is exactly similar to 
Case (a) and leads to exactly the same conclusions. It is 
left as an exercise to be worked out by the student. 

Case (c). When the forces are equal and opposite^ but 
not acting along the same line, they form a couple. These 
will be treated later. (See Art. 30.) 

Problem 9. Two parallel forces, one of 20 lb. and one of 
100 lb., have lines of action 24 in. apart. Find the resultant in mag- 
nitude, direction, and point of application : 

(1) When they are in the same direction. 

(2) When they are in opposite directions. 

Problem 10. A horizontal beam of length I is supported at its 
ends by two piers and loaded with a single load P at a distance of 

T from one end. Find the pressure of the piers against the beam. 

Problem 11. The locomotive shown in Fig. 19 is run upon a 
turntable whose length is 100 ft. Find the position of the engine 
so that the table will balance. 

21. System of Parallel Forces in Space. — If the forces 
are all parallel, it is evident that the resultant is equal in 
magnitude to the algebraic sum of the forces, and that its 
line of action is parallel to the forces. It remains, then, 
to determine the point of application of this resultant. 



I 



PARALLEL FORCES 



23 




Fig. 19 



Suppose the forces represented by Pj, P^, P3, P4, etc., and 
let their points of application be ^i^i^i, ^2^/2^2' ^zVz^z'^ 
x^y^^, etc. (Fig. 20). (In order to avoid a complicated 
figure only two forces are shown.) The two forces 




Fig. 20 



24 APPLIED MECHANICS FOR ENGINEERS 

Pj and P2 li^ i^^ ^ plane and have a resultant R = P^ + P^ 
whose point of application is at a distance z^ from the xy- 
plane and on a line joining L^ and L^ at L^ , Draw L^A 
perpendicular to L^A, Then from Art. 20, R^ L^B = 
P^L^A^ which multiplied by sec a gives R L^L^ = P^LJLy^ 

L,y P. 

or —2 — = 1 

-^2^1 -^1 + ^2 

Now in the plane of z^ and z^ draw ig^ ^^^ -^'^' perpen- 
dicular to 2-^, and we have 



z^ — Zc 



^2^1 -^1^ ^1 ~ % 



so that 2' - ^2 = p ^ p (^1 - ^2)' 

-^1 + ^2 

A + ^2 

Consider now ^' with P^\ these forces lie in a plane. 
Let their resultant be R^^ and its point of application 
x^\y^^^z^\ Following out the above reasoning for this 
case, the resultant is seen to be R^^ = P^ + P^ + P^ and 

z'^ = -^ ^ "^ -^3^3 __ -^1^1 "I" ^2^2 "^ -^3^3 . 

Extending this process so as to include all of the forces 
P4, Pg, etc., and calling the final resultant M and its point 
of application x, y, i, we have 

R= P^+P^ + P^ + Pi+ etc. 

and i - -Pi^i+-P2^2 + A ^3 + -P4^4 + etc. ^ 2^ 
Pi + P^ + A + A + etc. 2P' 



PABALLEL FORCES 25 

and by a reasoning similar to the above 

y = -Pi.Vi + A .y2 + Aj/« + ^ 4?/ 4 + etc. ^ 2Pz/ 
^ P^ + F^ + P^ + F^ + etc. SP 

_ _ PiX^ + P<>x<2, + Pg^s + -^A + etc. _ 2Px 

This point of application of the resultant is called the cen- 
ter of the system of parallel forces. 

As an illustration of the above, suppose P^ = 50 lb., P^ 
= 100 lb., ^3=300 lb., P4 = 10 1b., andP5 = -400 lb., 
and their points of application respectively 2, 1,— 5; — 1, 
-2,4; 2,1,-2; -2,1,1; 1,1,1. The resultant in this 
case equals 50 lb. + 100 lb. + 300 lb. + 10 lb. - 400 lb. = 
60 lb. and its point of application 

__ 50 (2) + 100 (-1) + 300 (2) + 10 (-2)- 400(1) _^ 
^- QQ -^. 

_ __ 50(1) + 100(- 2)+ 300(1) + 10(1)- 400(1) _ 

^ ~ 60 "" "" ' 

__50( _5) + 100(4) + 300(-2) + 10(l)-400(l) _ 

"^ - 60 - "■ ^■^• 

As another illustration consider the problem of finding 
the center of the system of parallel forces P^, P^^ Pg, in 
Fig. 21. The figure represents a Z-iron of the same cross 
section throughout, and Pj, P^-, and Pg are therefore the 
weights of the individual parts (considering the Z-iron as 
divided into three parts — two legs and the connecting ver- 
tical portion). If the weight of a cubic inch of iron =.26 
lb., Pj = .78 lb., P2 = 2.08 lb., P3 = 1.04 lb., and therefore 
i2 = 3.9 lb. The points of application of P^, P^^ and Pg 



26 



APPLIED MECHANICS FOR ENGINEERS 



are (- ^, - 1 91), Q, - 1-, 5), and (2, - i, i), respectively, 
so that 



_ .78(- 1)4- 2.08(1) + 1.04(2) _ 
'^~" 3.9 



= .70 in., 



i/ 



78(-i) + 2.08(-i) + 1.04(--i) ^ ^g^.^^^ 



3.9 



.78(-V-)+ 2.08(5) + 1.04(1) _ 
3.9 "■ 



m, 



This point x^ y, 2J is, in this case, the center of gravity of 
the Z-iron. 

Problem 12. Parallel forces of Pp Pg' ^s' ^^^ ^4 ^^^ ^^ the corners 
of a rectangle 3 ft. by 2 ft. and perpendicular to its plane. Find the 
point of application of the resultant, if P^ = 10 lb., Pg = ^0 lb., Pg = 

100 lb., P^ = 200 lb., Pj and P^ being 
2 ft. apart, and Pg on same side as Pg. 

Problem 13. Eight parallel forces 
act at the corners of a one-inch cube, 
making an angle of 45° with one of its 
faces. Find the point of application 
of the resultant force, if Pj = 30 lb., 
Pg = 50 lb., P3 = 10 lb., P, = 20 lb., 
P, = 100 lb., Pq=5 lb., P, = 10 lb., 
Pg = 40 lb. 

The student should prove that 
the moment of the resultant of 
any system of parallel forces in 
space with respect to any line 
in space, equals the sum of the moments of the forces with 
respect to this same line. The solution of Problems 12 
and 13 is made much shorter by using this principle. 




Fig. 21 



CHAPTER IV 



CENTER OF GRAVITY 



22. Definition of the Center of Gravity. — The center of 
gravity of a body may be defined as the point of appli- 
cation of the resultant attraction of the earth for that 
body, and the center of gravity of several bodies con- 
y sidered together, as the point of application of 
the resultant attraction of the earth for the 
bodies. The expressions for x, y, and z. Art. 
21, may be used for locating the center of 
gravity, in the latter case, without change, P^, 
P^, Pg, etc., 'representing the weights of the 
individual bodies. In such cases the center of 
the system of parallel forces is the center of gravity 
of the body. The attention of the stu- 
dent is called to the fact that the forces 
acting upon the particles of a body, due 
to the attraction of the earth, are not 
parallel, but meet in the center of the 
earth. For all practical purposes, how- 
ever, they are considered 
parallel. 



^1 



B 







F„ 







Fig. 22 

27 



28 APPLIED MECHANICS FOR ENGINEERS 

If the unit weight times the volume be substituted for 
weight, that is, if we write instead of P^, 7^ V^ and P^^ 
72 ^"2, etc., then S, y^ z become 

- ^ 7i^"i ^1 + 72 ^^2 ^2 + 78 ^s^3 + etc. ^ ^7^^ 
7il^i + 72^'2 + 73f'^3+etc. 27r' 

— __ 27 Vy - __ 27Fi 

^ ■" 27 r' ^ " 27 r* 

And if the bodies are all of the same material and so 
have the same heaviness, 7 is constant and may be taken 
outside the summation sign, where it cancels out. This 
gives values for S, y, and i, 

^_2F^ 7/-^ZM 5-2Z^ 

formulae exactly similar to those of Art. 21, where the P's 
are replaced by F's. 

If the bodies are thin plates of the same material, of 
constant thickness 5, we may write for F^, V^, F3, etc., 
hF^, hF^, bF^, etc., where the P's represent the areas of 
the faces of the plates. Making this substitution for the 
F's, 5, ^, z may be written 

- _ IF^x^ + hF^x^^ + hF^ X r^ + etc. ^ ^Fx 
^■" bF^ + bF^ + bF^ + etc. "" 2P' 



- 2P7/ 2P2: 



the J being a constant factor, cancels out. These formulae 
are applicable to finding the center of gravity of areas, and 
are much used by engineers for finding the center of 
gravity of sections of angles, channels, T-sections, Z-sec- 



CENTER OF GRAVIir 



29 



^~ 








~" 


/ 


j_0.4" 


[ 










*;^ 




^ 


^. 




^ 






1 


10 


' 



Fig. 23 



tions, etc. As 
an illustration 
let it be re- 
quired to find 
the center of 
gravity of the 

angle section shown in Fig. 22. It is convenient to select 
the X- and ^/-axes as shown and to divide the area up into 
the two indicated areas F^ and F^. We then have 

F, + F^ ^ F, + F, ' 

x^T/^ being the center of gravity of F^, and 0:2^2 the center 
of gravity of F^. It is left to the student to make the 
numerical substitution and to calculate the values for x 
and ^. 

Second Method. The same results for x and y might be 
obtained from the expressions 



X 



- -^1^1 - F^H .-7 _ F^yi - F^y^ 



F,-F, 



y 



F,-F, 



where now F^ is the area formed by completing the rec- 
tangle whose sides are 4 in. and 3 in. and x^y^ the center of 
gravity of this rectangle referred to the coordinate axes, 
and F^ the area of the rectangle whose sides are 3 in. and 
2 in. and x^y^ the coordinates of its center of gravity re- 
ferred to the same axes. Let the student find the center 
of gravity of the angle section by this method and com- 
pare the results with those obtained by the previous 
method. 



30 



APPLIED MECHANICS FOR ENGINEEES 



Problem 14. Find the center of gravity of the channel section 
shown in Fig. 23. 

Problem 15. Find the center of gravity of the T-section shown 
in Fig. 24. 



'' .. 1 
. f I 




Fig. 24 



Fig. 25 



Problem 16. Find the center of gravity of the U-section shown 

in Fig. 25. Given the fact that the center of gravity of a semicircular 

4r 
area is — from the diameter. (See Prob. 22.) 

Problem 17. Find the position of the center of gravity of a 
trapezoidal area, the lengths of whose parallel sides are a^ and Og? re- 
^ ^ spectively, and the distance 

between them h. (See Fig. 
26.) 

Hint. Draw the diaiio- 
nal AB and call tlie tri- 
angle ACB, F^, and the 
triangle ABD, F^. Given, 
the center of gravity of a triangle is \ the distance from the base to 
the vertex. (See Prob. 21.) Select ^jD as the a:-axis, then 




Fig. 26 



y 



^l.Vl + ^2.^2 



where y^ = | A and ^2 = i ^* ^^^ center of gravity is seen to lie on a 
line joining the middle points of the parallel sides. 



CENTER OF GRAVITY 



31 



"^- Problem 18. A cylindrical piece of cast iron whose height is 
6 in. and the radius of whose base is 2 in., has a cylindrical hole of 1 in. 
radius drilled in one end, the axis of which 
coincides wdth the axis of the cylinder. The 
hole was originally 3 in. deep, but has been 
filled with lead until it is only 1 in. deep. 
Find the center of gravity of the body, the unit 
weight of lead being 710 and of cast iron 450. 
(Fig. 27.) 

Problem 19. Find the center of gravity of 
a portion of a reinforced concrete beam. (See Fig. 27 

Fig. 28.) The beam is reinforced with three half-inch steel rods, 
centers 1 in. from the bottom of the beam and 1 in. from the sides. 
The center of the middle rod is 4 in. from the sides. 
(y for steel = 490 lb. per cubic foot ; 
y for concrete = 125 lb. per cubic foot.) 
Note. It is seen that the thickness cancels out of the expression 
for the center of gravity, and might, therefore, have been neglected. 

Z 




^ 







yf'^ 



<^--6^-\6^ 



Iz: 



Fig. 28 



32 APPLIED MECHANICS FOR ENGINEERS 

23. Center of Gravity determined by Symmetry. — In 
some areas and solids it is often possible to determine the 
center of gravity from considerations of the symmetry of 
the figure ; for example, the center of gravity of a paral- 
lelogram is at its geometrical center. This is also true of 
the circle, square, cylinder, sphere, etc. Whenever any 
axis is an axis of symmetry, that is, an axis such that for 
every element of area or volume on one side there is an 
equal area or volume on the other side, symmetrically 
placed, the center of gravity must be on that axis. This 
was found to be true in the case of the channel section, 
the T-section, and the U -section. In each of these cases 
the vertical line through the center of gravity is an axis 
of symmetry. The student will be able to note many 
more such cases, and by a little thought will often be 
able to find either x^ y^ or z from observation. 

24. Center of Gravity determined by Aid of the Calculus. 

— The expressions for x^ y^ and z used to locate tlie center 
of gravity in Art. 22 may be put in the form of the quo- 
tient of two integrals, and these expressions may be inte- 
grated when there is no discontinuity in the expressions 
between the limits taken. With this understanding we 

may write __2Px_|^P(f) 

__ ^Pz _^dP{z) 
" 2P ~ {dP 



CENTER OF GRAVITY 



83 



As an illustration, suppose it is desired to obtain the 

center of gravity of a right circular cone of altitude h and 

radius of base r. Take the 2:-axis as the axis of the cone 

with the vertex at the origin. (See Fig. 29.) It is evident 

z 




Fig. 29 



that ^ = and 2 = 0, so that it is only necessary to find x. 

The volume of any dv cut from the cone by two parallel 

planes, perpendicular to x and separated by a distance 

dx, is iry^dx^ and the weight of this dv is ^iryHx = dP. 

Therefore ^ ^n 

I xdP 1 x^iry^dx 



x = 



I dP I ^iry^dx 



But from similar triangles y : x 
gives 2 

sc = 



r : h or y = -x. 



This 



12 Jo 



x^dx 



A' 



y^T-oJ xHx T 



r 

X2 



r 
3 



3 , 
h 4 



34 



APPLIED MECHANICS FOB ENGINEERS 



The expressions for x, y, and 2, involving c?P, may be 
changed to similar ones involving dv, and these become 
for homogeneous bodies, since dP = ydv^ 

1 xdv j ydv j zdv 

J dv I dv j dv 

and for thin plates of constant thickness b the dv may be 
replaced by hdF, giving values of x, y, and 5 for area, 

^xdF _^ JydF __ jzdF 
7 ' ^=-p . ^ = -7 



X 




Fig. 30 

The center of gravity of thin homogeneous wires of con- 
stant cross section may be found by replacing the dv in 
the above formulae by ads, where a is the constant area 
of cross section and ds is a distance along the curve. The 
formulae then become 



J 



xds 



ds 



[yds 






CENTER OF GRAVITY 



35 



Problem 20. Find the center of gravity of a parabolic area shown 
in Fig. 30, the equation of the parabola being i/ = 2px. 



I 



xdF 



J 



dF 






Here dF = ydx, so that 

^ — — a ^ ~~ " 2 ^n** ^ 

Ci/dx V2^CJdx ^^^]o 

Jo Jo 

It is left as a problem for the student to show that y = ^ h. 

Y 




Fig. 31 



Problem 21. Find the center of gravity of a triangle whose alti- 
tude is h and whose base is a. Take the origin at the vertex and 
draw the 2:-axis perpendicular to the base. (See Fig. 31.) 






. Here dF = (y -\- y')dx, and from similar triangles 



?/ + V' rr ??x, so that dF = - xdx, 
^ ^ h h 



36 



APPLIED MECHANICS FOR ENGINEERS 



and 



0? = 



hjo"^ 3 Jo 




■2 ~h 



h. 



?H- -I 
hJo 2_\ 



The center of gravity is f the distance from the vertex to the base, 
and since the median is a line of symmetry, it is a point on the median. 
It is, in fact, the point where the medians of the triangle intersect. 

Problem 22. Find the center of gravity of a section of a flat 
ring, outside radius R^ and inside radius i^g. (See Fig. 32.) Let 
the angle of the sector be 2 0. Take the origin at the center and let 
the X-axis bisect the angle 2 6. 



a 



xdF 



i 



dF 



here, dF = pdpda, and x = p cos a. 



so that 



\ i cos a da ' p'^dp 
r (pdpda 




Fig. 32 

Integrating the numerator first, 

R 



i cos ada { p^df 



Rl_ 



3 



3 /^ + 



R 8 

cos ada — — ^ 



^(2sin0) 



CENTER OF GRAVITY 

Integrating the denominator, 

da I pdp = ^'J^ — ^ J ./(^ = (/?^2 _ /^g^) (9. 

Therefore, 



37 






If i?2 = 0, the sector becomes the sector of a circle, and x becomes 



2 ^ sin e 
3 



^ = ^^1-^ 



If the sector is a semicircle, that is, if 2 5 = ir, then, since = ^, 

- 2p fll 




Fig. 33 



Problem 23. Find the center of gravity of a portion of circular 
wire (Fig. 33) of length L and whose chord = 2h. Take the center 
of the circular arc as origin and let the x-axis bisect Z. Then 

( xds 
X = -^ ; but R\x — ds\ dy, 

ds= -dy, 

X 



'B 



R I dy 

'ft _ 2 ii6 _ radius x chord 

I ~ L " arc 



38 



APPLIED MECHANICS FOB ENGINEERS 



For a semicircular wire 



X — 



Diameter 



TT 



Problem 24. Find the center of gravity of a paraboloid of revo- 
lution. The equation of the generating curve being z/^ = 2 jox, and 
the greatest value of x^ is a. 

Note. Use the same method as that used for the right circular 
cone. 

r 




Fig. ai 



Problem 25. Find the center of gravity of a semi-ellipse (Fig. 34) 
whose equation is 






X — 



J 



xdF 



s 



dF 



where 



dF=2ydx = 2- Va^ - x^ dx, 



CENTER OF GRAVITY 



39 



therefore 



ciJo _ 

X = —. 7^^ 



1 f""** 

8^ ^ 



^'Va-^c/x i[.Va-.'^ + a^sin-g; 



_ 3 _4a 

¥ * 2 

Problem 26. Find the center of gravity of a hemisphere, the 
radius of the sphere being r. Let the equation of the generating 
circle of the surface be x^ -^ 7/ = r^. Then 

CxdP 
X =^^ , where dP=yTry'^dx = y7r(r^ - x^)dx^ 



J 



dP 



and ^^y^(r^ - a:2)rfx = yw\rx - g-J,,^-^— 



X = 



JyTrr' 



= ?r. 



Problem 27. Find the center of gravity of 
the area between the parabola, the y-axis, and the 
line AB in Problem 20. 

Problem 28. A quadrant of a circle is taken 
from a square whose sides equal the radius of tlie 
circle. (See Fig. 35.) Find the center of gravity 
of the remaining area. Fig. 35 

Problem 29. Suppose that the corners A and C of the angle iron 
in Fig. 22 are cut to tlie arc of a circle of y\^ in. radius and the angle 
at B is filled to the arc of a circle of | in. radius ; what would be the 
change in x and ^? 




40 APPLIED MECHANICS FOR ENGINEERS 

K 

Problem 30. Show that the center of gravity of the segment of a 
circle (Fig. 36), included between the arc 2 5 and the chord 2dy is 

given by ^ = — — , where F is the area of the segment. 
12 F 




Fig. 36 

25. Center of Gravity of Counterbalance of Locomotive Drive 

Wheel. — In Fig. 37 the drive wheel is indicated by the 
circle and the counterbalance by the portion inclosed by 
the heavy lines, the point is the center of the wheel, and 
a is the angle subtended by the counterbalance. The 




point 0' is the center of the circle forming the inner 
boundary of the counterbalance, and yS is the angle sub- 
tended by the counterbalance at this point. Let F^ repre- 
sent the area of the segment of radius r and F^ the area 



CENTER OF GRAVITY 41 

« 

of the segment of radius r^ Also let x-^ represent tlie 

distance of the center of gravity of F^ from 0, and x^ the 

distance of the center of gravity of F^ from 0^ Then, 

from Problem 30, „ „ o q 

8 a^ i / 8 a*^ 
x-^— and x^' = 



12F^ ' 12 F^ 

But 2^2, the distance of the center of gravity of F^ from (?, 

So that a:, the distance of the center of gravity of the 

F X ~ F X 

counterbalance from (9, equals —^^^ — ^^^~^' ^^ ^^ seen that 

F^ equals area of sector minus area of triangle equals 
ar cos -, and similarly. 



i<„ = ^—-^ — ar. cos ^. 

2 .> 1 w7 



/3 
2^ I r^ coSy-— r cos 

Therefore, a; = — 



26. Simpson's Rule. — When the algebraic equation of a 

curve is known, it is expressed as y =/(2:), and the area 

between the curve and either axis i^ always determined 

by integration. In Fig. 38 the area ABCD is expressed 

by the integral ^^ ^^ 

jjjdx = jj(x)dx, 

when the curve represented by y —f(x) is continuous 
between A and B. 

In many engineering problems the curve is such that 
its equation is not known, so that approximate methods of 



42 APPLIED MECHANIC^ FOR ENGINEERS 

obtaining the areas under the curve must be resorted to. 
One of these methods of approximation is known as 
Simpson s Rule, Suppose the curve in question is the 

r 




Fig. 38 

curve AB (Fig. 39) and it is desired to find the area 
between the portion AB and the a;-axis. Divide the 
length I — a into an even number of equal parts n (here 
71=10). Consider the portion OBUF 2ind imagine it mag- 
nified as shown in Fig. 40, Pass a parabolic arc through 















n D 










^R 


c ^3^ 
















_^ 


A 


[^ 






















/" 


^ 


y. 


f, 


V, 


E 


2/4 


2/5 
F 


2/6 


y-i 


Vs 


Vo 


2/10 




n 


7 






\JV 




















^ 















Fig. 39 



the points 0, D, Gr\ then the area CDEF is approxi- 
mated by the area of the parabolic segment CGrDI plus 
the area of tlie trapezoid CDEF, therefore area GGrBFF 
= K2/2 + yd^^+ 1(2/3 - 2 [^2 + ^4])^^' since the area of 



CENTER OF GRAVITY 



43 



the parabolic segment is | tlie area of the circumscribing 
parallelogram. Since EII= = A a:, this area may be 



71 



written 



lA<2/2 + 4^/3 + 2/4)- 



In a similar way the next two strips to the right will have 
-— — (y^ + 4 y. + ^g), and the next two strips, an 



an area. 



A y 

area, (2/6 + "^ 2/7 +2/3)' ^^^ ^^ ^^^- Adding all these so 

o 

as to get the total area under the portion of the curve AB^ 
we get 



total area 



h — a 
3 . 10 



[2/0 + Kvi + y^ + yr, + yi + y^) 



+ -0j2 + y^ + yQ + ys) + yio] > 




ys 



D 



or in general for n divisions, 

total area = -^ [^/o + -^C^i + 2/3 + ^5 + * * * ^n-i) 
o n 

+ -(2/2 + ^4 + 2/6 + ••• 2/^-2) + 2/n]. 

and this is Simpson's formula for determining approxi- 
mately the area under a curve. It is easy to see that 
the smaller A2:, the less the approxi- 
mation will be. 



27. Application of Simpson's Rule. — 

Simpson's Rule may be made use of 
in determining approximately not only 
areas, but volumes and moments. On 
account of its use in adding moments 
Simpson's formula may be employed in 
finding the center of gravity of areas 
or volumes bounded by lines or sur- 
faces whose equations are not known. 
Suppose, for example, it is desired to 



E 



n 

Fig. 40 



44 



APPLIED MECHANICS FOR ENGINEERS 



know the volume and position of the center of gravity of 
a coal bunker of a ship as shown in Fig. 41. The bunker 
is 80 ft. long and the areas A^, A^, A^, A^, A^, are as 
follows: ^, = 400 sq.ft., 

^^ = 700 sq. ft., 

^2=650sq. ft., 

^3=600 sq. ft., 

A^ = 400 sq. ft. 

The distance between the successive areas is 20 ft. 
Applying Simpson's formula for volume, 

Qf) 

Summing the values A^Xq, A^x^, A^x^, etc., we obtain 



'S,vx = 



80 



(3) (4) 

A, 



lA^o + 4(^12^1 + Agx^} + 2 A^x^ + A^x^\, 

A^ A^ A^ At 



^ 


y^ y\ ^i 

X\ X: X\ 


y 




L^ 


1 

1 

1 

1 
1 
1 

1 

1 

1 
1 

-—4-— 

/ _ 


...->- 


1 
1 
1 

1 
1 
1 


J 



Fig. 41 



where x^ = 0, a^j = 20, x^ = 40, x<^ = 60, x^ = 80. The po- 
sition of the center of gravity from the fore end can now 
be obtained from the relation 

^vx 



x — 



CENTER OF GRAVITY 



45 



A value of x might also have been obtained from the 
formula 

Vq + ^'l 4- v^ + ^3 + ^4 

by simply adding the terms in the numerator and denom- 
inator. Compare the value obtained 
by using this formula with that ob- 
tained by using Simpson's formula. 

Problem 31. A reservoir with five- 
foot contour lines is shown in Fig. 42. 
Find the volume of water and the distance 
of the center of gravity from the surface 
of the water, if the areas of the contour 
lines are as follows : Aq = 0, A^ = 100 sq. 
ft., A, = 200 sq. ft., A, = 500 sq. ft., A, = ^^^ ^^ 

600 sq. ft., .45 = 1000 sq. ft., A^ = 1500 sq. 

ft., Aj = 2000 sq. ft., Ag = 2500 sq. ft. Making substitutions in the 
Simpson's formula, it becomes, for the volume, 

Yo\nme = -^lA,-\.i(A, + A,-\-A,+ A,)^2(A,-^A,-\-A,)+A,-]. 

Summing the values AqXq^ A^x^, ^2^2* ^^-j by Simpson's formula, 
we have 




'2(VX — 



^ 



(3) (8) 



IAqXq-^4:(A^x^-\-A^x^-\-A^^x^-{-AjX.) 



4- 2(A2X2 + A^x^ + A^^)+A^x^'], 



where Xq =0 ft., x^ = o ft., arg = 10 ft., etc., so that 

V 

Both numerator and denominator are computed by Simpson's formula. 
Compute X by means of the formula, 

- _ V(^X(^ + v^x^ + vnX2 4- ^V^s 4- etc. 

1^0+^1 + ^2 + ^3+ ^^' 

and compare with previous result. 

Problem 32. Compute x for the parabolic area of Fig. 30, by 



46 



APPLIED MECHANICS FOB ENGINEERS 



using Simpson's Rule, and compare the result with that obtained by- 
integration. 

Problem 33. By Simpson's Rule find the area and center of 

gravity of the rail section shown 




in Fig. 43. 

All = 2.05 
^10 = 2.07 



A,= 



A^ =1.89 



A.= 



.55 

.61 

1.95 



A^=A9 

A, = m 

^5 = .51 
A, = .ol 

and the horizontal distances are 
as follows : 



A, = 



.82 



^0 = 2.95 



= 4'' 



Uc 



1.24' 



u. 



1.0'' 



= 4.08" w-. = 1.18" 
= 4.24" w. = 1.0" 



= 2.5" 



u, = 1.0" 



Wg = 1.24" 

M2 = 2.23" 
u^ = 5.5" 
u, =6" 



Fig. 43 

Problem 34. Find the center of gravity of the deck beam section 
shown in Fig. 44. Use the formula 

— _ F^xi + F2X2 + F's^s + ^'^^• 
Fi + F "+ F, + etc. ' 




and divide the bulb area into convenient 
areas, say F^, F^, F-,, etc. Check the re- 
sult thus obtained with that obtained by 
balancing a stiff paper model over a knife 
edge. 

28. DuranJ's Rule. — A method 
of finding the area of irregular 
areas was published by Professor 
Durand in the Engmeering Neivs^ 
Jan. 18, 1894. The rule states 
that the total area of an irregular 
curve equals 




Fig. 44 



CENTER OF GRAVITY 



47 



(^^)[(1 - 0-f>)^^0 + (1 + 0.1)^1 + ^2 + ^'3 + ^'4 + % 

+ ...(i+o.iK_i + (i-o.6X], 



.^« 



where the i^'s have the same 
meaning as before. The divi- 
sions may be even or odd. The 
student is advised to make use 
of this rule as well as Simpson's 
Rule and compare the results. 




Fig. 45 



j,tt," 29. Theorems of Pappus and 
Guldinus. — Let the disk in Fig. 
45 be any slice cut from a solid 
of revolution by two parallel 
planes perpendi-eular to the axis 

of revolution and at a distance dx apart. The volume of this 
slice is dv = 7r(i/^^ — y^^dx^ so that the volume of the whole 

solid is V == IT i Qy^ — y^^dx. The generating figure of 

this slice, dF^ equals {^y^ — y^dx^ and the distance of its 

center of gravity from the ^^-axis is ^2_^1. We have 



seen that 



then 



_ CydF 



and this, considering the expression for volume, becomes 

1 



y = - 



therefore, 



V 



2irF 
yF. 



48 APPLIED MECHANICS FOR ENGINEERS 

This may be stated as a general principle as follows : 
The volume of any solid of revolution is equal to the area 

of the generating figure times the distance its center of 

gravity moves. 

Problem 35. Find the volume of a sphere, radius r, by the above 
method, assuming it to be generated by a semicircular area revolving 
about a diameter. 

Problem 36. Assuming the volume of the sphere known, find the 
center of gravity of the generating semicircular area. 

Problem 37. Find the volume of a right circular cone, assuming 
that the generating triangle has a base r and altitude h. 

Problem 38. Assuming the volume of the cone known, find the 
center of gravity of the generating triangle. 

Problem 39. The parabolic area of Problem 20 revolves about 
the a;-axis ; find the volume of the resulting solid. 

Problem 40. Find the vol- 
ume of an anchor ring, if the 
radius of the generating figure is 
a and the distance of its center 
from the axis of revolution is r. 

Let the curve AB (Fig. 
46), of length Z, be the gen- 
^^* crating curve of a surface 

of revolution. The area of the surface generated by ds will 
be dF= 2 nryds,, and the area of the whole surface will be 
F = 2 7r i yds. The center of gravity of this curve AB is 
given by the expression 




= jfyds; 



Cyds 

jds 

- F 

y =- — -., ov F^t-Kyl- 
lirl 



CENTER OF GRAVITY 49 

This may be stated as follows : The area of any surface of 
revolution is equal to the length of the generating curve times 
the distance its center of gravity moves. 

Problem 41. Find the surface of a sphere, radius r, assuming the 
generating line to be a semicircular arc. 

Problem 42. Find the center of gravity of a quadrant of a circu- 
lar wire, radius of the circle r; use results obtained above. 

Problem 43. Find the surface of the paraboloid in Problem 39. 



CHAPTER V 



COUPLES 



30. Couples Defined. — In Art. 20, Case (c), it was shown 
that the resultant of two parallel forces in a plane was 
equal to the algebraic sum of the two forces. The con- 
sideration of the case when the forces are equal and oppo- 
site in direction, that is, where the resultant is zero, will 



^^ 



^i^ 



{a) 




A 



Pi 
Fig, 47 



ib) 



now be considered. It is easy to see that since the result- 
ant is zero, the two forces tend to produce only a rotation 
of the rigid body about a gravity axis perpendicular to the 
plane of the forces. Such a pair of equal and opposite 
parallel forces is called a couple. Let it be represented as 
in Fig. 47 (a), the two forces being P, and d the distance 
between the lines of action of the forces. This distance 
d is called the arm of the couple ; one of the forces times 

50 



COUPLES 51 

the arm is called the moment of the couple. It was found 
ill Art. 20 that the algebraic sum of the moments of the 
resultant and the system of parallel forces with respect to 
any point in their plane, is zero. In this case, since the 
resultant is zero, the moment of the forces of the couple 
with respect to any point in the plane is equal to the sum 
of the moments of the two forces with respect to that 
point. Let the point be (7, Fig. 47 (a), distant x from 
the force P; then — Px — P(^— d — x) represents the sum 
of the moments of the two forces with respect to the point 
C (calling distance below O negative). This sum is equal 
to Pc?, the moment of the couple. The student should 
take C in different positions and show that the moment of 
the two forces with respect to any point in the plane is 
always Pd, Since the moment consists of force times 
distance, it is measured in terms of the units of force and 
distance ; that is, foot-pounds or inch-pounds, usually. If 
the couple tends to produce rotation in the clockwise 
direction, the moment is said to be negative; and if counter- 
clockwise, positive. 

31. Representation of Couples. — The couple involves mag- 
nitude (moment) and direction (rotation), and may, there- 
fore, be represented by an arrow, the length of tlie line 
being proportional to the moment of the couple, and the 
arrow indicating the direction of rotation. In order to 
make the matter of direction of rotation clear, the agree- 
ment is made that the arrow be drawn perpendicular to 
the plane of the couple on that side from which the rota- 
tion appears counter-clockwise. This means that if we 
look along the arrow pointing toward us, the rotation 



52 



APPLIED MECHANICS FOR ENGINEERS 



the couple 
any point 



appears counter-clockwise. Thus, the couple of Fig. 47 
(a), whose moment is Pc?, may be represented b}^ the arrow 
in Fig. 48 (a), where the length of line AB is propor- 
tional to Pd and the couple is in a plane through B and 
perpendicular to AB. The line AB is sometimes called 
the axis of the couple; it may be drawn perpendicular 

to the plane 
of 
at 

in that plane, 
since the mo- 
ment is con- 
stant for any 
point in the 
plane. In a 
similar way, 
the couple (i) 

in Fig. 47 whose moment is Bid^ is represented completely 
by the arrow (5), Fig. 48, the length CD being propor- 
tional to F^dy 

IN'oTE. The arrows are placed slightly away from the ends, so that 
the moment arrows may not be confused with force arrows. These 
arrows, like force arrows, may be added algebraically when parallel, 
resolved into components and compounded into resultants ; the prin- 
ciple of transmissibility holds and also the triangle and polygon laws 
as seen for force arrows. Several important conclusions follow easily 
as a result of this arrow representation. 

Since a moment arrow represents both force and distance and direc- 
tion of rotation, it is evident that it cannot be balanced by a single 
force arrow even though they have the same line of action and are 
opposite in direction. Hence, we conclude that a single force cannot 
balance a couple. 





Fig. 48 



COUPLES 



63 



32. Couples in One Plane. — If the couples are all in the 
same plane, their moment arrows are all parallel, and may- 
be added algebraically, so that the resultant couple lies in 
the same plane and its moment is the alyehraic surn of the 
moments of the individual couples. 

For example, in Fig. 49, the couples P^d^, ^2^2' ^3^3' 
P^d^, P^d^, P^d^^ are all in the plane (aJ) ; their resultant 
couple must also be in this 
plane, and its moment must be 
equal to the algebraic sum of 
the moments of these couples. 

It is evident since the above 
is true that a couple may be 
transferred to any part of its 
plane without changing its 
effect upon the rigid body 
upon which it acts. This 
means, when applied to some 
particular rigid body, as a 
closed book, that the effect of 
a couple acting in the plane of one of the covers of the 
book (book remains closed) tends to produce rotation 
about an axis, perpendicular to the cover through the 
center of gravity of the book ; and that this rotation 
is the same no matter where the couple acts, provided it 
remains always in the same plane. The moment arrow 
of the resultant couple will be perpendicular to the cover 
of the book and on the side from which the rotation 
appears counter-clockwise. The student should endeavor 
to see the application of the above theorem and to see that 
it agrees with his observations. 




Fig. 49 



54 APPLIED MECHANICS FOR ENGINEERS 

33. Couples in Parallel Planes. — The moment arrow 
represents a couple in magnitude and direction of rotation 
and shows that the plane of the couple is perpendicular to 
its line. This moment arrow represents any couple of 
given moment and direction in any plane perpendicular to 
its line. It is evident, then, that a couple may he trans- 
ferred to any parallel plane zvithout changifig its effect upon 
the rigid body upon which it acts. Applied to the case of 
the book in the preceding article, it may be said that the 
effect of the couple would be unchanged if it acted in the 
plane of the other cover or in any of the leaves. 

34. Couples in Intersecting Planes. — Suppose all the 
couples in a plane (1, 2) be added and let AB (a) (Fig. 
48) represent the moment arrow of the resultant couple, 
and let the sum of all the couples in the plane (2, 3) in- 
tersecting (1, 2) be represented by CD (J) (Fig. 60). 

These moment ar- 

c >— i> 

{()) ^ rows are perpen- 

^ dicular to their 
respective planes 
and may be moved 
about without 
changing the ef- 
fect of the couple. 
Move A and to some point on the line of intersection 
of the two planes. The resultant moment arrow is now 
found by the parallelogram law. The resultant couple 
has a moment represented by AU and acts in a plane 
perpendicular to AU and making an angle a with the 
plane (2, 3). 





COUPLES 55 

Problem 44. Two forces, each equal 10 lb., act in a vertical plane 
so as to form a positive couple. The distance between the forces is 
2 ft. ; another couple whose moment is equal to 20 in. -lb. acts in a 
horizontal plane and is negative. Required the resultant couple, its 
plane, and direction of rotation. 

Problem 45. A couple whose moment is 10 ft. -lb. acts in the 
a:?/-plane ; another couple whose moment is —30 in. -lb. acts in the 
a:2;-plane, and another couj^le whose moment is — 25 ft. ili-lb. acts in 
the ?/2;-plaiie. Required the amount, direction, and location of the 
resultant couple that w^ill hold these couples in equilibrium, (x, y, 
and 2-axes are at right angles with each other in this case.) 

/ 



CHAPTER VI 

NON-CONCURRENT FORCES 

35. Forces in a Plane. — The most general case of 
forces in a plane is that one in which the forces are non- 
concurrent and non-parallel. We shall now consider such 
a case. Let the forces be Pj, P^, Pg, P^, etc., as shown in 

T 




Fig. 51 



Fig. 51, and let them have the directions shown. For 
the sake of analysis, introduce at the origin two equal 
and opposite forces Pj, parallel to Pj, two equal and oppo- 



66 



NON-CONCURRENT FORCES 57 

site forces Pg' Parallel to P^^ and so on for each force. 
The introduction of these equal and opposite forces at 
the origin cannot change the state of motion of tlie rigid 
body. 

The original force P^ taken with one of the forces P^, 
introduced at the origin, forms a couple wliose moment is 
P^dy The same is also true of the forces P^^ Pg, P^, 
etc., giving respectively moments P2<^2' ^s^h' P^d^^ etc. 
In addition to tliese couples there is a system of con- 
curring forces at the origin Pj, P^, Pg, P^, etc. The 
resultant of this system is, as has been shown (Art. 16), 

R = V(S2:)2 + (S2/)2. 

The moments Pid^ -^2^2' ^3^3' ^^^-^ being all in one 
plane, may be added algebraically (see Art. 32), giving 
the moment of the resultant couple as SPc?. 

The systeyyi of non-concurrent forces in a plane may he 
reduced^ then^ to a single force R at the origin {arhitrarily 
selected^ and a single couple whose plane is the plane of the 
forces. 

For equilibrium, JK = and SfVZ = 0, or Sac = 0, ^y = 0, 
and SJPtZ = 0; that is, for equilibrium, the sum of the com- 
ponents of the forces along each of the two- axes is zero 
and the sum of the moments with respect to any point in 
the plane is zero. 

Considering as a special case the case Where the forces 
are concurring, it is seen that Pd is always zero (see 
Art. 16). The case of a system of parallel forces in a 
plane may also be considered as a special case of the above. 
(Art. 20 and Art. 31). 

Problem 46. The following forces act upon a rigid body : a 



58 



APPLIED MECHANICS FOB ENGINEERS 



1 TON 



2 TONS 



force of 100 lb. whose line of action makes an angle of 45° with the 
horizontal, and whose distance from an arbitrarily selected origin is 

2 ft. ; also a force of 50 
lb. whose line of action 
makes an angle of 120° 
with the horizontal, and 
whose distance from 
the origin is 3 ft.; and 
a force of 500 lb. whose 
line of action makes an 
angle of 300° wdth the 
horizontal and whose 
Find the resultant force and the 




Fig. 52 



distance from the origin is 6 ft. 
resultant couple. 

Problem 47. It is required to find the stress in the members ^5, 
EC, CD, and CE of the bridge truss showm in Fig. 52. 

^OTE. The member AB is the member between A and B, the 
member CD is the member between C and D, etc. This is a type of 
Warren bridge truss. All pieces (members) are pin-connected so 
that only tw-o forces act on each member. The members are, there- 
fore, under simple tension or compression ; that is, in each member 
the forces act along the piece. Usually, in such cases there are no 
loads on the upper pins. 

Solution of Problem. The reactions of the supports are found 
by considering all the external forces acting on the truss. Taking 
moments about the left support, we get the reaction at the right sup- 
port, equal 4500 lb. Summing the vertical forces or taking moments 
about the right-hand support, the reaction at 
the left-hand supj^ort is found to be 3500 lb. 

Cutting the truss along xy and putting in the 
forces exerted by the left-hand portion, consider 
the right-hand portion (see Fig. 53). The 
forces C and T act along the pieces, forming a 
system of concurring forces. For equilibrium, 
then, 2x = and 2^/ = 0, giving two equations, 
sufficient to determine the unknowns C and T. The forces in the 
members CD and CE may now be considered known. 




4600 LB&. I 

Fig. 53 



NON-CONCUBBENT FOBCES 



59 



Cutting the truss along the line ZW and putting in the forces 
exerted by the remaining portion of the truss, we have the portion 
represented in Fig. 54. This 
gives a system of concurring 
forces of which C and 2 T are 
known, so that from the equa- 
tions ^x = and 2?/=0 the 
remaining forces d and e may 
be found. 




Fig. 54 
The truss is pin-con- 



2 TONS 



Problem 48. Find the stress 

in each of the members AB and 

BC of the simple roof truss shown in Fig. 55. 

nected, and all the members are under simple tension or compression ex- 
cept the horizontal 
piece, which is under 
flexure. The method 
of cutting the truss, 
employed in the 
preceding problem, 
cannot be employed 
to advantage here, 
since the stress in 
the horizontal piece 
is not along the 

piece. The simplest method of solution for such a case is to take 

the whole member in question and consider all of the forces acting. 

In this case we have (see Fig. 

56) a system of non-concur- 
ring forces in a plane. For 

equilibrium 2^ = and 2?/ = 

and SPrZ = 0, from which 

P, 

mined, 




Fig. 55 



^1 

i 



P, and Pj may be deter- 



T 



f 



1 TON 



1 TON 



Fig. 56 



Problem 49. In the crane shown in Fig. 57 (a) find the forces 
acting on the pins and the tension in the tie AC. Tlie method of 
cutting cannot be used in this case since the vertical and horizontal 



60 



APPLIED MECHANICS FOB ENGINEERS 



members are in flexure. Taking the horizontal member and consid- 
ering all of the forces acting upon it, we have the system of non-con- 
curring forces shown in Fig. 
57 (6). Three unknowns are 
involved, Pg, Pi, and P2, and 
these may be determined by 
three equations ^x = 0, 2?/ 
== 0, and ^Pd = 0. It is to 
be remembered that the pin 
pressure at E is unknown in 
magnitude and direction. In 
all such cases it is usually 
more convenient to resolve 
this unknown pressure into 
its vertical and horizontal 
components, giving two un- 
known forces in known di- 
rections instead of one un- 
known force in an unknown 
direction. This will be done 
in all problems given here. 
4 TONS In the present case the two 
forces Pi and P2 are the 
components of the unknown pin pressure. 

The tension in the tie A C may be found by considering the forces 
acting on the whole crane and taking moments about B. Thus 
^Pd = gives, calling the tension in the tie T, 

T35sin45° = 8000 (25), 
y^ SOOO (25) 




Fig. 57 



or 



35 sin 45' 



Problem 50. In the crane shown in Fig. 58 (a) find the tension 
in the ties T and T' and the comj^ression in the boom. The method 
of cutting may be used here to determine the tension T and the com- 
pression in the boom, since AB is not in flexure, if we neglect its own 
weight. Cutting the structure about the point A and drawing the 
forces acting on the body, we have the system shown in Fig. 58 b. 



NON-CONCURRENT FORCES 



61 




c/ (b) 



Fig. 58 



The forces T^ may be considered as acting at the center of the pulley. 
The system of forces is concurring, so that 2x == and 2^ = are suf- 
ficient to determine T and C. 
T may be found by consider- ^ -^*v, 
ing the forces acting on the 
whole crane and taking mo- 
ments about the lowest point 
B, 

Note. Neglecting friction, 
the tension W in tlie cord sup- 
porting the weight is trans- 
mitted undiminished through- 
out its length. 

Problem 51. Find the 
horizontal and vertical com- 
ponents of the forces acting 
on the pins of the structure shown in Fig. 59. 

Suggestion. First take the vertical strip and consider all the 

forces acting on it. 

Problem 52. Find the forces 
acting on the pins of the structure 
shown in Fig. 60, the weight of the 
members AD, BF, and CE being 
600 lb., 400 lb., and 100 lb., respec- 
tively. 

Problem 53. A traction engine 
is passing over a bridge, and when 
it is in the position shown in Fig. 
61 one half of the load is carried 
by each truss. The weight of the 
engine is transmitted by the floor 
beams to the cross beams, and 
these are carried at the pin connections of the truss. Find the stress 
in the members AB^ BC, CE, CD, and DF, for the position of the 
engine shown. 




Fig 



62 



APPLIED MECHANICS FOR ENGINEERS 



XoTE. The floor beams are supposed to extend only from one 
cross beam to another. 

Problem 54. In Problem 50, suppose the weight of the boom to 

be one ton ; find the tensions 
T and T and the pin pres- 
sures. 

Note. The boom is now 
under flexure, so that the 
method of cutting cannot be 
made use of. 

Problem 55. A dredge 
or steam shovel, shown in 
outline in Fig. 62, has a dip- 
per with capacity of 10 tons. 
When the boom and dipper 
are in the position shown, find the forces acting on AB, CD, and EF. 
Suggestion. Consider first all the forces acting on CD, then all 
the forces acting on AB. 




Fig. 60 




Fig. 61 

Note. The member EF has been introduced as such for the sake 
of analysis ; it replaces two legs, forming an A frame. The projection 
of the point i^ is 6 ft. from the point E, 

Problem 56. Suppose the members of the structure in Problem 



NON-CONCURRENT FORCES 



63 



55 to have weights as follows: AB, 15 tons, and CD, 3 tons, not in- 
cluding the 10 tons of dipper and load. Find the stresses as required 
in preceding problem. 




Fig, 62 

Problem 57. Suppose the beam in Problem 5 to be 20 ft. long 
and to have a weight of 2000 lb. ; find the pin reaction and the ten- 
sion in the tie. 

Problem 58. Assume that the compression members of the War- 
ren bridge truss of Problem 47 have each a weight of 500 lb. ; find 
the stress in the members jBC and CE. 

36. Forces in Space, Non-intersecting and Non-parallel. — 
If a rigid body is acted upon by any system of forces in 
space Pj, P^') P3, P4, P5, Pg, etc., making angles with 
the arbitrarily chosen axes, x^ y, and ^, a^, ^j, 7^ ; a^^ ySg, 72 ; 
ttg, ^3, 73, etc. (see Fig. 63), it can be shown that the 
system may he replaced hy a single force and a single couple. 
The single force acts at the origin, and its direction 
angles are a, /8, and 7. The resultant couple acts in a 
plane whose direction angles are X, /i, v. 

Introduce at the origin two equal and opposite forces 
Pj parallel to the line of action of P^. One of these 
forces Pj together with the original P^ form a couple 
whose moment is P^d^^ and this couple may be represented 
by a moment arrow at the origin, perpendicular to the 



64 



APPLIED MECHANICS FOR ENGINEEBS 



plane of the couple (see Art. 31). We thus have re- 
placed the force P^ by an equal and parallel force at the 
origin and a couple represented at the origin by a mo- 
ment arrow P-^dy Proceeding in the same way with P^, 
Pg, P4, etc., we finally have instead of the original system 




Fig. 63 

of non-concurring, non-parallel forces in space, a sys- 
tem of concurring forces P^ P^, P31 P^. etc., at the origin 
and a system of moment arrows P^d-^, ^2^2' -^3^3' ^^^"> 
represented at the origin. The forces may be combined 
into a single resultant as in Art. 17, and we then have 



NON-CONCURRENT FORCES 



66 



whose direction cosines are 

cos «= --, COS /3 = -j^, cos 7 = -^- 

Since the moment arrows also follow the same laws as 
the force arrows, we may also write tlie moment of the 
resultant couple, 



where M^ = Pi^i cos \ + P^d.^ cos ^2 + etc., 

3Iy = Pi'^i cos fi^ + P^d^ cos /i2 + etc., 
M^ = Pid^ cos v^ 4- -^2^2 ^^^ ^2 + e^e- 

The direction angles of ilfare X, /i, i^, and these are de- 
fined as follows (see Fig. 64) : 



cos \ = —~, 
M 



My 

COS M = ^ 



COS V = — ^ 

il!f 



This system of forces pro- 
duces a translation of the 
body in the direction of 
R and a rotation about a 
gravity axis parallel to M. 
li B=0 and ilf^O, the 
body only rotates or is 
translated with uniform 
motion, and if M= and 
H^ 0^ the body has only 
translation with possibly 
uniform rotation. For 
equilibrium both Ii= 
and M= ; that is, 2x = 
0, % = 0, 2^ = 0, ilf, = 0, 




Fig. 64 



66 APPLIED MECHANICS FOB ENGINEERS 

My = 0, and M^ =0, or expressed in words: the sum of the 
components of the forces along each of the three arbitrarily 
chosen axes is zero^ and the sum of the moments with respect 
to each of these axes is zero. 

It may be further shown that the single force and 
resultant of the preceding article may be replaced by a 
single force and a couple whose plane of rotation is per- 
pendicular to the line of action of the force. 

Suppose iHfand R both drawn at the origin and let a be 
the angle between them. M may be resolved into com- 
ponents along and perpendicular to i2, Gr cos a along jR, 
and Gr sin a perpendicular to R, Gr sin a may be replaced 
by another couple having the same moment. Let the 
forces he — R and + R and allow the — R to act along 
the line of action of the resultant force. The other force 
of the couple acts along a line parallel to the direction of 
the resultant force. The forces + R and — R along the 
line of action of the resultant force neutralize each other, 
and we have left (a) a force, R^ parallel to the original 
resultant, and (5) a couple, Gr cos cc, acting in a plane per- 
pendicular to R. 

That is, the system reduces to a single force and a sin- 
gle couple whose plane is perpendicular to the line of 
action of the force, or we may say, the effect of any system 
of forces acting on a rigid body,, at any instant,, is to cause 
an angular acceleration about the instantaneous axis of rota- 
tion and an acceleration of translation along that axis. 
Such a system of forces is called a screw wrench. The 
instanta^^us axis is called the central axis. This axis 
passes thr<fWgh the center of gravity of the body if the 
body is free^ rotate. 



NON-CONCURRENT FORCES 



67 



Problem 59. A vertical shaft is acted upon by the belt pressures 
Tj and T^, the crank pin pressures ]\ 
and the reactions of the supports. See 
Fig. 65. Write down the six equations 
for equilibrium. 

Note. The 3/-axis has been chosen 
parallel to the force P, and T^ and Tg 
are parallel to the x-axis. 

^x = x' -{- x" - T^- T^ = 0, 

^z=z" - G - G' = 0] 

M,= -Pb- y"l = 0, 

My = x"l - T^c - T^c - G,'^= 0, 

Mz = Fa+ T^r - T^r = 0. 

From these six equations six unknown 
quantities can be found. If G^ Cg, 
Tp and Tg are known, the reaction of 
the supports and P may be found. 

Problem 60. A crane shown in 
Fig. 66 has a boom 45 ft. long and a 
mast 30 ft. high. It is loaded with 20 
tons, and the angle between the boom 
and mast is 45°. The two stiff legs 
each make angles of 30° with the mast and an angle of 90° with each 
other. Find the pin pressures in boom and mast, also the stress in 





Fig. GG 



68 APPLIED MECHANICS FOR ENGINEERS 

the legs when (a) the plane of the crane bisects the angle between 
the legs, and (b) the plane of the crane makes an angle of 30° with 
one of them. If the boom weighs 4000 lb., find the stress in the 
legs when the plane of the crane bisects the angle between them. 
Assume that the pulleys A and B are at the ends of the boom and 
mast respectively. 

Problem 61. Suppose the shaft of Problem 59 to be horizontal, 
find P and the reactions of the supports. Assume y horizontal and 
perpendicular to the shaft, and x vertical. 



CHAPTER VII 

MOMENT OF INERTIA 

37. Definition of Moment of Inertia. — The study of 
many problems considered in mechanics brings to our at- 
tention the value of the integral of the form (y^dF^ where 
dF represents an area and y the distance of the center of 
gravity of that area from an axis of reference. A more 
general definition of moment of inertia would be the prod- 
uct of an elementary area, mass^ or volume by the square of 
its distance from a designated pointy line^ or plane. The 
integral given above simply adds these products to give 
the moment of inertia of an entire area. In the case of a 
mass, the integral becomes jy^dM, and of volume (y^dV. 
If the area, mass, or volume is not continuous throughout, 
the limits of integration must be properly taken to 
account for the discontinuity. We shall designate 
moment of inertia by the letter /. Thus we write: 

I=jy^dF, 
I=^yHM, 

I=fy^dV, 

for area, mass, and volume, respectively. In finding the 
moment of inertia of several disconnected parts, it is often 



70 APPLIED MECHANICS FOR ENGINEEBS 

necessary to use the summation sign instead of the inte- 
gral ; we may then write : 

1= ^fdF, 
1= ^i/dM, 

for area, mass, and volume, respectively. 

Many problems that confront the engineer involve in 
their solution the consideration of the moment of inertia. 
This is the case when the energy of a rotating fly wheel, 
for example, is being determined. The energy of a rotat- 
ing body (kinetic energy, Art. 133) is expressed as follows: 

Kinetic energy = — -, 

where i"is the moment of inertia with respect to the axis 
of rotation and co is the angular velocity (see Art. 95). 
It is seen that the energy of rotating bodies, having the 
same angular velocity, or the same speed, is directly pro- 
portional to their moments of inertia. The quantity, 
therefore, plays a very important part in the considera- 
tion of rotating bodies. 

In computing the strength of a beam or column it is 
necessary to consider the product of an area and the 
square of its distance from some line. This is called 
the moment of inertia of the area; it is usually taken 
with respect to a gravity axis of the area. An idea of the 
use of moment of inertia in computing the strength of 
beams may be obtained by considering the beams sup- 
ported at both ends and loaded in the middle. The load 
that the beam will carry is then given by the formula 



MOMENT OF INERTIA 71 

P = — ^, where P is the load in pounds, I the length of 
Ih 

the beam in inches, h the height of beam in inches, p the 
strength of the material, of wliich the be.im is comj)osed, 
in pounds per square inch, and I is tlie moment of inertia 
of the cross section with respect to a horizontal gravity- 
axis. The student should thoroughly master the princi- 
ples of moment of inertia. 

38. Meaning of the Term. — The term moment of inertia 
is somewhat misleading, and the student is apt to try 
to connect moment of inertia with inertia. The term 
has no such significance and should be regarded as 
the name arbitrarily applied to a quantity that engineers 
frequently use. 

Radius of Gyration. The moment of inertia of an area 
involves area times the square of a distance. We may 
write 1= iy'^dF^FW^ where F is the area and Z: is a 
distance, at which, if the area were all concentrated, the 
moment of inertia would be unchanged. This distance 
k is called the radius of gyration. In a similar way 
for a mass we w^rite: /= \y\iM=Mk^^ and for volume 
1= ^y\lV= Vk^. 

39. Units of Moment of Inertia. — The moment of inertia 
of an area with respect to any axis may be expressed 
as Fk^. The area involves square inches, and ^ is a dis- 
tance squared. The product is expressed as inches to the 
fourth power. The moment of inertia of a volume Vk^ 
requires inches to the fifth power. Tlie moment of 
inertia of a mass requires in addition to Vk'^ the factor ^, 



72 APPLIED MECHANICS FOR ENGINEERS 

SO that, pounds and feet per second are involved. This is 
somewhat more complicated since it involves units of 
weight, distance, and time. This presence of ^(=32.2) 
in the expression requires that all distances be in feet. It 
is customary to express the moment of inertia of a mass 
without designating the units used, it being understood 
that feet, pounds, and seconds were used. 

40. Representation of Moment of Inertia. — From the 
definition of moment of inertia it is evident that an area 
has a different moment of inertia for every line in its 
plane. We shall designate the moment of inertia with 
respect to a line through the center of gravity by Ig with 
a subscript to indicate the particular gravity line in- 
tended. For example, Ig^. indicates the moment of inertia 
with respect to a gravity axis parallel to x^ and Ig^ indi- 
cates the moment of inertia with respect to a gravity 
axis parallel to y. The moment of inertia with respect 
to a line other than a gravity line will be designated 
by P, the proper subscript indicating the particular 
line. Similar subscripts will be used to designate the 
corresponding radii of gyration. It should be noted that 
moment of inertia is not a quantity involving direction. 
It has to do only with magnitude and is essentially 
positive. 

41. Moment of Inertia ; Parallel Axes. — Consider the 
area inclosed by the irregular line (Fig. 67) and sup- 
pose that its moment of inertia is known with respect 
to every line through its center of gravity (gravity 
axis). It is required to find its moment of inertia 
with respect to any other line in the plane. Select the 



MOMENT OF INERTIA 



73 




Fig. 67 



arbitrary rectangular axes x and y and draw through 
the center of gravity of the area a line parallel to the 
a;-axis. Let the 
distance between 
these parallel 
lines be d. The 
moment of inertia 
of the area i^ with 
respect to the 
dotted line will 
be called Ig^ and 
its moment of 
inertia with re- 
spect to the a;-axis, J'^. The moment of inertia of the 
elementary area dF with respect to the 2:-axis is then 
written (c? + d^^dF^ and the moment of inertia of the 
whole area F with respect to this same axis becomes 

r^= j(d + dydF= ^d?dF+ J2 dd^dF+ JQdydF. 

But i d^dF may be written d^jdF=d^F, since cZ is a 

constant, and 2d Cd^dF=2dQFd'} (see Art. 22), where 

d' is the distance of the center of gravity of the area 
from the line of reference. In this case d' = 0, so that 

2dCd^dF= 0. The term C{dydF equals Ig^ by defini- 
tion (see Art. 37). It follows, then, that 

or, expressed in words. 

The moment of inertia of an area with respect to any line 
in its plane is equal to its moment of inertia with respect to 



74 



APPLIED MECHANICS FOE ENGINEERS 



a parallel gravity axis plus the area times the square of the 
distance between the ttvo axes. 

This theorem is used very often in work that follows 
and should be thoroughly understood. It may also be 

by transposing terms. In this form it is convenient when 
J'^, jP, and d are known, and Ig^ is to be determined. It 
is easily seen from either of these expressions that the 
moment of inertia of an area for a gravity axis is less 
than for any other line in the plane. 

In case neither axis is a gravity axis, 2 dd^F is not 
equal to zero and T^= I' -\- Fd?' + 2 dd'F^ where I' is the 
moment of inertia with respect to the parallel axis and dJ 
the distance of the center of gravity of the area from this 
parallel axis ; the quantities F^ c?, and F ^ having the same 
meaning as before. 

42. Moment of Inertia ; Inclined Axis. — It is often de- 
sirable, when F^ is known, to find the moment of inertia 

Y 




MOMENT OF INERTIA 75 

with respect to an axis tv making an angle a with x (see 

Fig. 68). Here, I\,=:Jv^dF'dnd r, = Jw\IF. In terms 
of a;, y, and a, 

T^= j (7/ cosa — X sin a)^ dF 

= j ^2 (>Qg2 adF— 2 i XT/ cos a sin a c?l'+ j a;^ sin^ a dF 
= cos^ a j ?/2c?j^ — 2 sin a cos a j xydF+ sin^ a j a;^^?^ 

= i'^ cos^ a — sin 2 a I xydF + Ty sin^ a. 

In a similar way 

7'^ = I^ sin2 a + 2 sin a cos a j a^yc^i^ + Fy cos^ a. 

These are the required formulae for obtaining the mo- 
ment of inertia with respect to inclined axes. It follows 
that Ti \ ji — jt I ji 

That is, the sum of the moments of inertia of an area 
with respect to two rectangular axes in its plane is the 
same as the sum of the moments of inertia with respect to 
any other two rectangular axes in the same plane and 
passing through the same point. This states that the 
sum of the moments of inertia for any two rectangular 
axes through a point is constant. It will be seen in 
Art. 45 that this constant is the polar moment of inertia. 

43. Product of Inertia. — The integral j xydF is called 

a product of inertia^ for want of a better name. In case 
the area has an axis of symmetry, either the x- or y-axis 
may be taken along such an axis. The product of inertia 
then becomes zero, since if x is the axis of symmetry. 



76 APPLIED MECHANICS FOR ENGINEERS 

for every + y there is a corresponding — ?/. A similar 
reasoning shows the product of inertia zero when 7/ is the 
axis of symmetry. In such cases 

I'^^ = I'x C0S2 a + I'y sin2 a 

and j'^ = l'^ sin^ a + I'y cos^ a. 

When j xydF is not equal to zero it is necessary to 

select the proper limits of integration and sum the 
integral over the area in question. This is illustrated in 
Article 63. 

44. Axes of Greatest and Least Moment of Inertia. — It is 

often important to know for what axis through the center 
of gravity the moment of inertia is least or greatest ; that 
is, what value of a makes /^ or /^ a maximum or a mini- 
mum. For any area i^, i^, and I xydF are constant after 

the X and y axes have been selected. Using the method 
of the calculus for finding maxima and minima, we have, 
putting 



/ 



xydF= z, --—■ = (/y — i^) sin 2 a — 2i cos 2 a, 



Equating the right-hand side to zero, the value of a that 
gives either a maximum or minimum is seen to be given 
by the equation 













9 


i 




^y 


■i: 


or, 


what 


is 


the 


same 


thing. 




sin 


2a = 






2i 


and cos 


2a = 



lu-I. 



± V4 t-2+ (/^_/j2 ± V4>+ (J,-/J2 



MOMENT OF INERTIA 



11 



It is seen upon substituting these values of sin 2 a and 
cos 2 a in 

^'^ = 2(7, - /^)cos 2 a + 4 i sin 2 a 

that the positive sign before the radical indicates a mini- 
mum and the negative sign a maximum value for i^. 
Investigating the values of a which give I^ maximum or 
minimum values, it is seen that the value of a for which 7^ 
is a minimum gives 7^ maximum, and the value of a for 
which i^ is a maximum gives I^ minimum. These axes 
for which the moment of inertia is greatest and least are 
known as the Principal Axes of the Area. This subject 
will be further discussed in Art. 53. 

It is seen from the above that when either the x- or 

2/-axis is a line of symmetry, so that I xi/dF= 0, the val- 
ues of a which give maximum or minimum values for 
7^ and 7y are all zero. 
This means that the x- and 
y-axes, themselves, are the 
principal axes. 

45. Polar Moment of In- 
ertia. — The moment of 
inertia of an area with 
respect to a line perpen- 
dicular to its plane is called 
the polar moment of inertia 
of the area. 

Consider the area repre- 
sented in Fig. 69 and let the axis be perpendicular to the 
area at its center of gravity. Let dF represent an infini- 




Fia. 69 



78 



APPLIED MECHANICS FOR ENGINEERS 



tesimal area and let r be its distance from the axis. 
Representing the polar moment of inertia by i^, we have 



but 
so that 



r^ = x^ + ^2, 



or 



That is, the polar moment of inertia of an area is equal 
to the sum of the moments of inertia of any tivo rectangular 
axes through the same point. It has already been shown 
that I^ + ly^ constant (see Art. 42) for any point of an 
area. 

46. Moment of Inertia of Rectangle. — Let it be required 
to find the moment of inertia of the rectangle shown in 
Fig. 70 (a), with respect to the axis, x. We may write 



/ 



;=/yW. 



Since dF=bdy^ this becomes 

4 = *TW^=^' 



w^ 



42/ 



-.Y 



{a) 




Fig. 70 



MOMENT OF INERTIA 79 

To find the moment of inertia with respect to a gravity- 
axis parallel to x it is only necessary to make use of the 
formula Ig^ = 7'^ — Fd^^ from which we have 



],^ = ^bMandk%. = ^ 



From comparison we may write the moment of inertia 
with respect to a gravity line perpendicular to x^ 



and the polar moment of inertia for the center of gravity 

47. Moment of Inertia of a Triangle. — It is required to 
find the moment of inertia of the triangle shown in Fig. 
70 (J) with respect to the axis a:, coinciding with the base 
of the triangle. We have 

j^ = j yHF^ where dF = xdy, 

But x = -(h — I/), from similar triangles, giving 
ft 

The moment of inertia with respect to horizontal gravity 

36 



axis may now be determined. 1^^=!^ ^—Fd?=-^^^ and 



"" "18" 



80 



APPLIED MECHANICS FOR ENGINEERS 



It is left as an exercise for the student to find tlie moment 
of inertia with respect to an axis through the vei'tex paral- 
lel to the base, and also the polar moment of inertia for 
the center of gravity. 

48. Moment of Inertia of a Circular Area. — The moment 
of inertia of a circular area with respect to a horizontal 
gravity axis a;, as shown in Fig. 71, may be found as fol- 
lows : J^^ = j 2/^dF, Changing to polar coordinates, re- 
membering that y = psm6^ and dF= dp(^pd6^^ the inte- 
gral becomes 



I,, = fp^ sin^ OpdOdp. 




Fig. 71 



This integral involves two variables, p and 0. It will, 
therefore, be necessary to make use of a double integra- 
tion. For this purpose, write 



'-gx 



sin2 edej pHp = -j 1 (1 - cos 2 e')de 

4L2 4 Jo ~ * 



MOMENT OF INERTIA 



81 



The corresponding radius of gyration is kg^ = ^. On 

account of the symmetry of the figure tliis is the moment 
of inertia for any line in the plane through the center of 
gravity. It follows that 



[l --^* 


[ ^.-"'■* 


OV 1 

and that < 


^ 2 




^P -2- 




Fig. 72 



49. Moment of Inertia of Elliptical Area. — Let it be 
required to find I^^ and I^y of the elliptical area shown 
in Fig. 72. The equation of the bounding curve is 



^ + ^=1 
a^ V' 



and 



I^y =^x\lF =fx^ 2 ydx. 



From the equation of the bounding curve 



^ a 



82 APPLIED MECHANICS FOB ENGINEERS 

SO that 



Iay = — I x^'V a? —x^ dx 
^y a Jo 

= —[^(2 x^ - a2) V^2T^ + ~ sin-i-T" 
a[_8 8 a Jo 

= — T"'^ ^^^ therefore kgy = g. 

In a similar way 

I^^=fy^dF=Jf2xdy 

= "/Jo^^^^^ - / ^2/ = ^^^ and therefore ^^^ = ^. 
Since /^ = 7^^, + J^^, the polar moment of inertia is 



ab 



TT 



^ (a2 + 62)^ and k^ = ^ Va2 + 52. 

It is seen that when a = b = r the equations obtained 
for the elliptical area are the same as those obtained for 
the circular area, just as they should be. 

50. Moment of Inertia of Angle Section. — When an area 
may be divided up into a number of triangles, or rec- 
tangles, or other simple divisions, the moment of inertia 
of the whole area with respect to any axis is equal to the 
sum of the moments of the individual parts. This method 
is often made use of in determining the moment of inertia 
of such areas as the section of the angle iron, shown in 
Fig. 73. 

We shall now determine the moment of inertia of this 
section with respect to the horizontal and vertical gravity 
axes, Ig^ and J^^, and also with respect to an axis v (see 



MOMENT OF INERTIA 



83 



Art. 53), making an angle a with the axis x. Consider 
the section divided into two rectangles, one 5" x f '' which 
we may call jF\ and the other 3|'' x f ^', which we may call 
Fc^. The moment of inertia of the section, with respect to 
X, is equal to the moment of inertia of F^ with respect to x 
plus the moment of inertia of F^ with respect to x, so that 
la. = M^Xiy + 5(|)(.808)2 + J,(2gl)3 1 + -V-(f )(1-19/ 
= 7.14 in. to the 4th power. Similarly 

I.y = iWi-Xiy + ¥(l)(l-30)^ + iV(f )(5)^ + 6(|)(.88)=' 
= 12.61 in. to the 4th power. 

Note. The problem of finding the moment of inertia of angle 
sections, channel sections, Z-bar sections, and the built-up sections 
shown in Figs. 75, 76, 77, 78, 79, is of special interest and importance 
to engineers, occurring as it does in the computation of the strength 
of all beams and columns. gy 













. — i- 


> 


n 


> 




i 


I .^- 






'^ 


-T y 


?4 

t 








. 












,/ 


i 


k: — 1.62 







^9^ 



9^ 



Fig. 73 



84 



APPLIED MECHANICS FOB ENGINEERS 



9^ 















' 


r -" 




i 


1 ^^ 


H\ 






- — 


—^>x-— 


— - 


-f""^ 


5 




.^-T 


»n 










c n 















gx 



9y Problem 62. Find 

the moment of inertia of 
the Z-bar section shown 
in Fig. 74 for the gravity- 
axes Qx and gy. 

Hint. Divide the area 
into three rectangles. 

Problem 63. Com- 
pute the moment of 
inertia for the channel 
I Fig. 74 section, shown in Fig. 23, 

Problem 14, for the horizontal and vertical gravity axes. 

Problem 64. Required the moment of inertia of the T-section (Fig. 
24, Problem 15), also the moment of inertia of the U-section (Fig. 25, 
Problem 16) with respect to both horizontal and vertical gravity axes. 

Problem 65. The section shown in Fig. 75 consists of a web 
section and 4 angles, as shown. Find the moment of inertia of the 
whole section with respect to the 
horizontal gravity axis. Given, 
the moment of inertia of an 
angle section with respect to its 
own gravity axis, g'^ is 28.15 in. 
to the 4th power. 

Problem 66. Consider the 
section given in Problem 65 to ^^^- '^^ 

be so taken that it includes two rivet holes, as indicated by the posi- 
tion of the rivets in Fig. 75. Compute the moment of inertia of the 
whole section, when the moment of inertia of the rivet holes is 
deducted. The distance from the center of the rivet hole to the out- 
side of the angle section may be taken as 3 in. Compare the result 
with that obtained in the succeeding problem. 

Problem 67. The same section shown in Fig. 75 is shown in Fig. 
76 with two cover plates. Find the moment of inertia of the whole 
beam section with respect to its horizontal gravity axis, now that the 
cover plates have been added. 




MOMENT OF INERTIA 



85 



Problem 68. Find the moment of inertia of the section of a box 
girder, shown in Fig. 77, with respect to its horizontal gravity axis. 
The moment of inertia of one of 
the angle sections with respect to 
its own horizontal gravity axis, is 
31.92 in. to the 4th power. 

Problem 69. Find the mo- 
ment of inertia of the column 
section, show^n in Fig. 78, with 
respect to the two gravity axes gjj 
and cjy. The column is built up ^^* ' 

of one central plate, two outside plates, and four Z-bars. The legs 
of the Z-bars are equal, and have a length of 3^ in. The moment of 

inertia of each Z-bar 







1.82' 



T 



-til 



g 



— ^fl" 



— g 



section with respect 
to its own horizontal 
and vertical gravity 
axis is 42.12 and 
15.44 in. to the 4th 
power, respectively. 

Problem 70. Find 

the moment of in- 
ertia of the section 
shown in Fig. 79, 
with respect to the 
horizontal and verti- 
cal gravity axes g-c 
and gy. This section 
is made up of plates 
and angles. The 
moment of inertia of each angle section with respect to both its own 
horizontal and vertical gravity axes is 28.15 in. to the 4th power. 

51. Moment of Inertia by Graphical Method. — It will 
often be necessary to lind the moment of inertia of a plane 
section whose bounding curve is of a complicated form, as 



-18- 



FiG. 77 



86 



APPLIED MECHANICS FOR ENGINEERS 



in the case when it is necessary to compute the strength 
of rails or deck beams. The graphical method given 
below may be used for such cases. 



:5 



S — EEEEH 



■*4 



ilt 



V>i* 



'V 

d 

,1J. 



xL 



9' 



tL 



9^ 



-16- 



99 



Fig. 78 



Let the section be a rail section, Fig. 80. Draw two 
lines, AB and CD, parallel to the required gravity axis, 
at any distance, Z, apart. The present section is symmet- 
rical with respect to the ^/-axis, so that it will only be 
necessary to consider the part on one side of that axis, say 
the part to the right. Suppose the section divided into 
strips parallel to AB and CD^ and let x denote the length 
of one of these strips, and dy its width. For each value 
of X there is a length x^ found, such that 

Then for every point P on the original section whose 
coordinates are x and y^ there will be a point P^ on the 
transformed section whose coordinates are x' and y. Sup- 
pose all these points, P\ constructed, and a boundary line 
drawn through them. Let J^ denote the area of the orig- 



MOMENT OF INERTIA 



87 



f/U 



^ 



uJi 



yiG. 



SM 



-6^v 



I X 



1.78" 



-^^—g' 



L 



Q'^ 



T 



Y I 



h28- 



^L^^Vl 



16 



Fig. 79 

inal curve, and F the area of the transformed curve, also 
N= ^ydF = ^yxdy = l^x'dy = IF. 

But \ydF=yF (^Art. 24), where j^ is the distance of the 
center of gravity of F from the line AB. It follows that 

This locates the center of gravity. 

The moment of inertia will be found by substituting 
for each x^ x\ and for each x\ x'\ such that 

Every point, P\ now goes over into a point P", forming 
a new transformed boundary. Call the area of this last 

y y y^ 

curve F^^ , Since x^^ = x^'-^ and x^ = x^^ x^' = x'^. 



88 



APPLIED MECHANICS FOR ENGINEERS 




Fig. 80 

Therefore the moment of inertia 

r^= ^fdF= ^y^xdy = pfx''dy=PF'^, 

giving the moment of inertia of the original section with 
respect to the line AB. 

To determine I^^ it is simply necessary to use the for- 
mula Ig-^ = I' ^ — Fd^, This gives 

The areas of the sections are measured by means of a 
planimeter. 

52. Moment of Inertia by Use of Simpson's Rule. — An 
approximate value for the moment of inertia of irregular 
sections, such as rail sections, may be obtained by the use 
of Simpson's Rule. Let the irregular area be the rail sec- 



MOMENT OF INERT I A 89 

tion (Fig. 43) and let it be required to find the moment 
of inertia of tlie section with respect to the base of the rail. 
We may write 

/; = -li/A = ylA, + y\A, + ylA^ - + y^A,, 

where the ^'s represent the areas and the y's the distance 
from the center of the ^'s to the base of the rail. In this 

case 2/0 == 4> Hi = 1' ^2 = 4' ^^^"> ^^^^ ^0 = --95, A^ = 1.95, 
7l2 = .61, etc. (see Problem 33). 

A more exact summation of the terms would be given 
by adding by means of Simpson's Formula. This gives 
(Art. 2G) 



r= 6 r 



-X 



3(12) 



+ 2(2/1^2 + i/4^'4 +•••) + yX I 



where u^, u-^^ ti^, etc., have the values given in Problem 33. 
The student should compare the result obtained by this 
method with tliat obtained by the method of direct addi- 
tion given above. Compare the value obtained with that 
resulting when Durand's Rule (Art. 28) is used. Use 
both methods to find the moment of inertia of the sections 
in Problem 33 and Problem 34. 

53. Least Moment of Inertia of Area. — In considering 
the strength of columns and struts it is necessary to know 
the least moment of inertia of a cross section, since bend- 
ing will take place about an axis of its cross section 
having such least moment of inertia. It was shown in 
Art. 42 that, if the moment of inertia of the area with 
respect to two rectangular axes in its plane is known, 
the moment of inertia with respect to any other axis. 



90 APPLIED MECHANICS FOR ENGINEERS 

making an angle a with one of these, could be found. It 
was further developed (Art. 44) that the value of a that 
would render the moment of inertia a minimum was 
given by the equation 



2 (xydF 
tan 2 a = — ^ —• 

In case either of the axes x or y is an axis symmetry, the 
value of a given by this criterion is zero, so that, for areas 
having an axis of symmetry, the axis of least moment of 
inertia is the axis of symmetry or the one perpendicular to it. 
As an illustration of the problem in general let it be 
required to find the least moment of inertia of the angle 
section shown in Fig. 73 with respect to any axis in the 
plane of the area through the center of gravity. Let v be 
the gravity axis making an angle a with the rr-axis. The 
problem then is to find such a value of a that Ig^ will be a 
minimum. From Art. 42 we have 

Igy = Ig^ cos^ cc " Sill ^ai xi/cixdy + Igy slu^ a. 

In Art. 50 it was found that Ig^ = 7.14 and Igy = 12.61. 

We proceed now to find the value of j xydxdy for the 

angle section. For this purpose, suppose the section com- 
posed of two rectangles, F^ (5 in. x | in.), and F^ (p^^ in. 
X I in.), and then find the value of the integral, for the 
two rectangles separately. Considering first the area F^^ 
and using the double integration, we get 

r-'' rJyLdy = P1T(zlM2I^ _ (^.495)^1 

•^1.62 *^-.495 ^1.62 L ^ 2 J 

= .505r^^M8)!_(l:62)2n 2.222. 



MOMENT OF INERTIA 91 

In a similar way for F^, we have 



.025 [^M2I^-I:995)_^1 = 3.288, 



2-J 

Therefore, j xydxdy for the whole area of the angle sec- 
tion is 5.51 in. to the 4th power. From this we find 

tan 2 a = ^1^ = 2.02. 
5.47 

Therefore 2 a = 63°40', 

a = 31° 50'. 

The expression for J^„ now becomes 
J,, = 7.14 cos2(31° 500 - 5.51 sin(63° 45') 

+ 12.61 sin2(31°500 = 3.72 in. to the 4th power. 

This gives the least radius of gyration, 

hg,= .84 in. 

Problem 71. Find the least moment of inertia Igy and least radius 
of gyration kg^ of the Z-section shown in Fig. 71. In this case 
Ig^ = 15.44 in. to the 4th power and Igy = 42.12 in. to the 4th power. 

Ans. Least Igy = 5.66 in. to 4th power and least kgy = .81 in. 

Problem 72. An angle iron has equal legs. The section, similar 
to that in Fig. 73, is 8 in. x 8 in. wdth a thickness of I in. Find 
Ig^y Igy, ^ast Igy aud least A:^. 

Ans. Ig^ = Igy =48.65 in. to the 4th power, 
Ig^ = 19.59 in. to the 4th power, 
kgy = 1.59 in. 

Problem 73. Find the moment of inertia of column section, 
shown in Fig. 78, with respect to an axis v making an angle of 30" 
with gx. What value of a gives 7^„ miiiinum in this case? 



92 APPLIED MECHANICS FOR ENGINEERS 

54. The Ellipse of Inertia. — It is interesting to note, at 
this point, the relations between the moments of inertia 
with respect to all the lines, in the plane of the area pass- 
ing through a point. We have seen that for every point 
in an area there is always a pair of rectangular axes for 
which the moment of inertia is a maximum or a minimum ; 
that is, there is always a pair of principal axes. The cri- 
terion for such axes was found to be 

tan 2a = __^^, 

which means, since the tangent of an angle may have any 
value from to infinity, positive and negative, that for 
every point there is always a pair of axes such that 



^ = 0, or j xydF= 0. 



This means that the expression for J^ may always be 
reduced to the form 

Jy = I^ cos^ (^ + ly sin2 a 

by properly selecting the axes of reference, where now 
Jp and ly represent the principal moments of inertia. If 
we divide through by F^ the equation becomes 

^2 -- ^2 cos^ a+k^ sin2 a. 
Let p=Mji, sothatyi;^ = ^and^^ = ^, 

fC^ Ky. K y 

then h\ = ^ — ^cos^ a + 2 — i: sin^ a, 

ky fc-^ 

or dividing by k% 

-j _ p2 cos^ ci .p^ sin2 a 



MOMENT OF INERTIA 93 

which is the equation of an ellipse referred to the princi- 
pal axes of inertia as axes. It may be written 

It is evident that k^ is inversely proportional to p, so 
that the major axis of tlie ellipse is along the axis of the 
least moment of inertia and minor axis along the axis of 
greatest moment of inertia. The ellipse of inertia has no 
physical significance, but merely shows the relation be- 
tween the moments of inertia with respect to the different 
axes through a point. If, then, the moments of inertia for 
all axes in a plane, through a point, be laid off on these 
axes, to scale, the locus of the end points will be an ellipse. 
The ellipse of inertia furnishes a graphical method for 
finding the moment of inertia for any axis through a point. 

55. Moment of Inertia of Thin Plates. — Suppose the 
plate of constant thickness t and unit weight 7 and let x 
be the distance of any dM from the axis of reference, then 

^x= j x^d3I= ^ I x\iV= - M x^dF. But this expression 

under the integral sign is the expression for the moment 
of inertia of the area of one of the faces of the plate, if F 
represents the area of a face. Therefore, the moment of 
inertia of a thin plate with reference to an axis in its j^^awe 

equals — times the moment of inertia of the area, of its face 

with reference to the same axis. 

A similar statement is seen to hold with reference to 
the polar moment of inertia of a thin plate by replacing x 
by p the distance of c?iirfrom a point. The following re- 
sults are deduced at once. 



94 APPLIED MECHANICS FOB ENGINEERS 



I, for thin circular plate : 

J _<ytf-7nf\_M^ y _ r 
""' A 2j- 2 ' ^^^~V2 



I, for thin rectangular plate : 



7„ = 2^aM.V^;*„ = 



VW+¥ 



9 



^''~ 2V3 



I^ for triangular plate : 

g\S6'-J 18 '••- 8V2 



Ig^ (parallel to base) = — f gg ^^^J = -j^ >Kx = — 



I, for elliptical plate 
(1) major axis : 



J- ^-/tfirab^^^Mlfi . j^ _h 



(2) minor axis : 

J- _jt/'Trb^\_Ma'^ J. _« 
"^"7^ 4 J T' "^"2 

(3) polar axis : 

■^ffp - ~ ~~^^ 4- « ; 1 — . f^pg - — 2 — 

(4) central axis, 2r, making angle a with major axis: 



^gv 



_-/t( 'rraW\ _Ma%^ j. ^ ab_ 
ff\ir')~ 4r^ ' "" 2r 



MOMENT OF INERTIA 95 

56. Moment of Inertia of Right Prism ; Geometrical 

Axis. — The moment of inertia of a right prism of height 
h with respect to its geometrical axis is given by the ex- 

That is, the moment of inertia of a rigid prism of height h 

is equal to — times the polar moment of inertia of its base. 
9 
From this result we may write : 

For right circular cylinder : 

Right parallelopiped : 

therefore, 

where d is the diagonal of the base. 
Hollow cylinder : 



r r^hfiTT\ 7rri\ M.^, ^. . Vr'^ + r'^ 



Elliptical cylinder : 



-^^2/-y-4-C^ +^)- 4 ' ^gy- 2 

57. Moment of Inertia of Right Prism ; Axis Perpendicu- 
lar to Geometrical Axis. — Let the axis be perpendicular to 
the geometrical axis through the base. Consider a tliin 
slice cut from the prism by two parallel planes, distant dy 



96 



APPLIED MECHANICS EOE ENGINEERS 



and perpendicular to the prism. The slice so cut may be 

considered a thin plate, Avhose mass is "^dvF^ where F is 

9 
the cross section of the prism. Suppose the distance of 

this slice from the base is y. Then the moment of inertia 

of this slice with respect to the axis through the base is 





{ci) Fig. 81 

equal to its moment of inertia about its own parallel 
gravity axis plus its mass times the square of the distance 
between the axes. (See Art. 41.) This will be made 
more clear by reference to the special case of the right 
circular cylinder of Fig. 81 (a). Adding the moments 
of all the slices, we liave 

g ''>g \ 3/ 

For the right circular cylinder : 



I. 



31 ('-J^ F=a/? 



A2 



MOMENT OF INERTIA 97 

For right elliptical cyliiider : 

(1) major axis : Fj, = 3ll y + — )• 

(2) minor axis: ![,= wf- + ^. 

For right rectangular cijlinder : 

(1) parallel to h: r, = 3lf^ + -\ 

(2) parallel to h, : 7j = iff (^^ + | 

58. Moment of Inertia of Solid of Revolution. — Con- 
sider tlie moment of inertia of a solid of revolution with 
respect to its axis of revolution. Imagine the solid cut 
into thin slices, all of same thickness, by parallel planes 
perpendicular to the axis of revolution. Each slice is a 
circular disk of thickness dg and radius re, and its polar 
moment of inertia with respect to the axis of revolution is 

^di/TTx^ • — • The moment of inertia of the solid of revo- 

^ . ^ 

lution is the sum of the moments of the small slices, so 
that 

the limits of integration and the relation between x and y 
depending upon the particular solid considered. 

For right circular cone : 

The right circular cone is illustrated in Fig. 81 (6). 
For this case 

"^ Jo 2a ^ 2fi ,Wo^ ^ lOrt 10 



f^^^cly. 



cj ■" 'Igh^^o" " 10 (/ 10 



since x = -y. 



98 APPLIED MECHANICS FOR ENGINEERS 

For a sphere : 

since a;^ = r^ — 7/^ 

= ^ ro^di/ - 2 T^dy + y'dy-) =Wy - ^ + 

= TZ: . A r^ = I Mr^ therefore A: 2 = f ^2. 
^ 15 5 ^'5 

i^or an ellipsoid of revolution : 
(1) prolate spheroid : 

2 a^-a z a a^^-a 



+r 



g^-a zg 



I 



since ^ = - v a^ — a;^, 

a 

and v = - irah\ L^ = -^^^^^ • -— a^ = | TMTft^. 

3 ga'^ 15 » 

(2) oblate spheroid : 

since ^ = t ^^^ ~" 3/^ ^^^ ^ = o ^r^^^^- 

6 3 

59. Moment of Inertia of Right Circular Cone. — When 
the moment of inertia of a right circular cone with respect 
to an axis through its vertex parallel to the base is to be 
found we may proceed as in Art. 58. 



MOMENT OF INERTIA 



99 



Imagine the cone to be cut by parallel planes into slices 
as shown in Fig. 81 (J). The moment of inertia of the 
whole cone with respect to x is equal to the sum of the 
moments of the small slices with respect to x. 



Then 






but 



x=--y 



h'' 



so that J^:=l!r!i^ + l!r?1^3^^/^3 2 + 3^2 
4^ 5 g b V20 . 



60. Moment of Inertia of Mass ; Parallel Axis. — If the 
moment of inertia of a mass with respect to an axis in 
space is known, it is im- 
portant to be able to 
determine its moment 
with respect to any par- 
allel axis. Let dM be 
the mass of a particle of 
the body, a' and a (Fig. 
82) the parallel axes dis- 
taiit cZ, and r' and r the 
distances of the mass 
from the two axes. 

From the triangle r^ = 
by dM and integrating over the body, we have 

^rHM=-^ ^r^HM+ ^dHM - J2 r^d cos edM, 




Fig. 82 



r'2 -f d'^— 2 r'd cos 6, multiplying 



100 



APPLIED MECHANICS FOR ENGINEERS 



But r' COS 0=d\ the distance of dM from a plane through 
a\ perpcndicuhxr to the plane a^ and a, 

so that ^r\UI= ^r^d3I+ d^M- 2 d^d'dM, 

or la = la' + d'^M - 2 d3Id^, 

where d^ represents the distance of the center of gravity 
of the mass from the plane through a' perpendicular to 
the plane of a' and a. 

From this relation, if the position of the body is known 
and its moment of inertia with respect to an axis in 
space, its moment of inertia with respect to any other 
parallel axis may be found. 

In particular, if d^ = 0, that is, if the center of gravity 
of the body lies in a plane through a^ perpendicular to 
the plane of a' and a, the relation reduces to 

That is, the moment of inertia of a mass with respect to any 
axis in space is equal to its moment of inertia ivith respect to 



^^ 



m 




mm^m 



Fig. 83 

a parallel axis, lyiyig in a gravity plane ^ perpendicular to the 
line joining the two axes^ pins the mass times the square of 
the distance between the bodies. 

The student will notice that this relation is very similar 
to the one developed for the moment of inertia of plane 
areas with respect to parallel axes, in Art. 41. 



MOMENT OF INERTIA 



101 



Problem 74. Show tliat the moment of inertia of a right circular 
cylinder, altitude h, and radius of base ?-, with respect tg a gravity 

axis parallel to the base is Igx = mI — h "/ ). J^^id find Wui moiucnt 



of inertia with respect to an axis parallel to this and at a distance d 
from the base. 

Problem 75. Show that the moment of inertia of a right circular 
cone, altitude h, and radius of base r, with respect to a gravity axis 
parallel to the base is Igx = ^^ M (r- + i A-). 

Problem 76. Find the moment of inertia of a slender rod of 
length I with respect to an axis through one end and perpen- 
dicular to the rod. Let the cross section be F and the mass M: then 

Problem 77. It is required to find the moment of inertia of the 
cast-iron disk fly wheel shown in Fig. 83 with respect to its geomet- 
rical axis. 

Hint. The wheel may be regarded as made up of three hollow 
cylinders, the moment of the whole wheel being equal to the sum 
of the moments of the three parts. The dimensions are as follows : 
diameter of wheel 2 ft., width of rim and hub 4 in., thickness of rim 
and web 2 in., thickness of hub 1\ in., and diameter of shaft 2 in. 
All distances must be in feet. 




'^ ss 



JL. 



Wm 



v3><- 



/i>6 



15 



Fig. 84 



102 APPLIED MECHANICS FOB ENGINEERS 

Problem 78. Find the moment of inertia of the cast-iron fly- 
wheel shown in Fig. 84 with respect to its axis of rotation. There 
are six elliptical spokes, and these may be regarded as of the same 
cross section throughout their entire length. 

61. Moment of Inertia of Non-homogeneous Bodies. — When 
the bodies are not homogeneous, the expressions for the 
moment of inertia given in this chapter do not hold, since 
in that case 7 is no longer constant. In case the law of 
variation of 7 is known, as for example if 7 varies as the 
distance from the line, then this variable value of 7 may 
be used and the moment of inertia found. 

Let it be required to find the moment of inertia of a 
right circular cylinder with respect to an axis through its 
base, if the density varies as the distance from the base, 
in such a way that 7 = 2/' where ?/ is a distance measured 
from the base (Art. 57). 

Then /;= Ci'^Fdykl + '^FdyyA 



=[ 



I4F y^ F /■ 



FJif^.oh . ¥ 
9 



= £±[T,f-^^'^ 



If 7 varies in some other way, the proper value must 
be used in the integral. In most cases, however, 7 is 
a constant. 

62. Moment of Inertia of a Mass ; Inclined Axis. — We 

shall now study the problem of finding the moment of 
inertia of a solid with respect to an axis inclined to the 
coordinate axes. Suppose the moments of inertia of the 



MOMENT OF INERTIA 



103 



body with respect to the three coordinate axes known 
from the expressions: 

and let it be required to find the moment of inertia of the 
body with respect to any other axis OA making angles 
a, yS, 7 with the coordinate axes. (See Fig. 85.) Let 
dM equal the mass of an infinitesimal portion of the body 
and d its distance from the axis OA. 

Since r^ = a;^ _|_ ^2 _(_ ^2^ q^ _ ^ ^^^ ol + y cos /3 + 2 cos 7 
and 

(p = 7^— OA^ = (x^+ ^2 -f 22) — (x cos a+ y cos /3 + 2J COS7)2, 

z 



dM 




t Fig. 85 

we may write 

= / [ (^ + ^^ + ^0 — (^ cos a + y cos /3 + 2 cos 7)^] dM, 



104 APPLIED MECHANICS FOR ENGINEERS 

This reduces, since cos^ a + cos^yS + cos^ 7 = 1^ to 
J^ , = j(f+ ^2) cos2 adM+ J (^2 + ^2) cos2 /3dM 

+ f (2j2 + 2/2) cos2 ydM— 2 cos a COS yS (xydM 

— 2 cos y8 cos ^KyzdM— 2 cos 7 cos a (xzdM, 
or 
io^i = -^ cos2 oc + J^/ cos2 ^ + Iz cos2 7 — 2 COS a COS ^ (xydM 

— 2 cos yS COS 7 r^/^c^iHf — 2 cos 7 cos a KxzdM^ 

which gives the moment of inertia of the body with 
respect to an inclined axis in terms of the moments of 
inertia with respect to the coordinate axes and the prod- 
ucts of inertia \xydM^ \yzdM^ 'diiAxxzdM. 

63. Principal Axes. — If the three products of inertia 
KxydM^ \yzdM^ and \xzdM are each equal to zero, the 
expression for Iq^ reduces to the form 

Iqa = Ix C0S2 Ci + ly C0s2 /3 + I^ C0s2 7. 

In this case the coordinate axes x^ y, and z are called 
the principal axes for the point and the moments i^, Z^, 
and I^ the principal moments of inertia. 

If the point is the center of gravity of the body and 
the products of inertia are each equal to zero, the prin- 
cipal axes are called the principal axes of the body. It 
can be shown that it is always possible to select the" 
coordinate axes x, y, and z so that the products of inertia 
given in the expression for Iqj^ will each be zero. It fol- 
lows that for every point of a body there exists a set of 
rectangular axes that are principal axes. 



MOMENT OF INERTIA 105 

Problem 79. Find the moment of inertia of the ellipsoid whose 
surface is given by the equation 

with respect to the axes a, i, and c, and with respect to an inclined 
axis making angles a, /3, y with a, h, and c, respectively. The volume 
of an ellipsoid is 

o o o 

and loA — la cos2 a + /j COS^ ^ -\- Ic cos^ y. 

64. Ellipsoid of Inertia. — It is always possible to re- 
duce the expression for Iqj^ to the form 

J^^ = I^ COS^ OL + ly cos^ 13 + I^ cos^ 7, 

by selecting the axes x^ y, and z so that the products of 
inertia are zero. 

Dividing this equation through by M^ we have 

^Ia = ^1 cos2 a + kl cos2 /3 + k^ cos^ 7. 
Let /) = ^f^% so that 



"'X 

so that 



Z. -^rzM. h — P^OA r.j.A -L ^ h"^OA 



^0 A^y ^^ ^^ ^x^y 



which when divided through by ^^^ becomes 
1 — (£_52?_^4. (p c^^-"^ yQ)^ Cp cos 7)' 



106 



APPLIED MECHANICS FOR ENGINEERS 



This is seen to be the equation of an ellipsoid whose 
semi-axes are k^ky^ kjc^^ and kjcy\ the equation may be 
written 



1 = 



X' 



k^ ky 



+ 



2/' 



2 7.2 



+ 



2 7_2 



f^x '^z '^x f^y 



where x\ y\ and z^ represent the coordinates of a point 
on the line OA at a distance p from 0. 

It is evident that if we draw all the lines through and 
then locate all points x^^ y^^ and z^ on these lines, such that 

p = ^ the locus of all the points will be the ellip- 

k 

soid of inertia for that point. For the position of the co- 
ordinate axes selected, the principal axes of the ellipsoid 
of inertia coincide with the principal axes of the body. 

Problem 80. Write the equation and construct the inertia ellip- 
soid for the center of gravity of a right circular cylinder, altitude h 

and radius r. 

Problem 81. Construct the 
inertia ellipsoid for the center 
of a solid sphere of radius r. 

Problem 82. Show that the 
moment of inertia of the seg- 
ment of the circle F\ (Fig. 86) 
with respect to the axis OZ is 

r-V^ + lsin4 

4\2 2 / 

the moment of the sector 
OBSA, minus J ah^, the mo- 
ment of the triangle OAB or 




Fig. 8(3 



^oz = Y^(2a-sin2aj, 



MOMENT OF INERTIA 107 

and the moment of inertia of Fi with respect to 05 is — ( - — - sin a ) , 
the moment of inertia of the sector, minus \ ha^, the moment of inertia 
of the triangle ^Oi? or Iqs = — Ta - sin a M + ^ sin^^ j ~| . 

Problem 83. Show that the moment of inertia of the counter- 
balance, Fig. 37, with respect to a line through 0, perpendicular to OS, 
and in the plane of the wheel, is 

Ioz=[f^{2a-sin2a) -!^^(^2fi-sin2l3y'-^^ - F,(00r,]^ 

where Fo = ^ ~ar. cos ^, 00' — r. cos ^ - r cos -, and t is the thick- 
2 2 2 2 

ness, as explained in Art. 25. 

Problem 84. Find the moment of inertia of the counterbalance. 
Fig. 37, with respect to a line through perpendicular to the plane of 
the wheel. It may be written 

/o = j y^ [4 a - sin 2 cc -2 sin a ^1 + ?sin2^]1 - ^ fd ^ - sin 2 )8- 
2 sin ^(1 + I sin^ |)] -f i^^^ - F,(00')4^^ 

f if 

65. Moment of Inertia of Locomotive Drive Wheel. — The 
drive wheel may be represented as in Fig. 87, and may be 
considered as made up of a tire, rim, twenty elliptical 
spokes, counterbalance, and equivalent weight on opposite 
side of center, and hub. The tire, rim, and hub may each 
be considered as hollow cylinders whose moments may 
be found as in Art. 56. The moment of the spokes is 
easily found by considering them elliptic cylinders (see 
Art. 57), with the short axis of the ellipse in the plane of 
the wheel. The moment of the counterbalance is equal 
to the moment of the weights carried by the crank pin, 
radius of counterbalance 



times 



radius of crank 



108 APPLIED MECHANICS FOR ENGINEERS 

The dimensions of the wheel are as follows : radius of 
tread 40'^, radius of inside of tire 36'^, width of tread 5'', 
outside radius of rim 36'', inside radius of rim 34'', width 
of rim 4y . There are twenty elliptical spokes 24" long, 
31" X 21". The hub is 10" outside radius, 4|" inside 
radius, and 8" thick, radius of crank pin circle 18". The 
counterbalance has an outside radius of 34", an inside 




Fig. 87 

radius of 7' 11.5", thickness 7^", mass 20.2, and distance 
of its center of gravity from the center 28.8", a = 94° 40', 
yS = 30° 20', 7 = 490 (Art. 64). 

We shall neglect the moment of inertia of the flange 
and shall consider the spokes cylindrical throughout 
their length, and that 10% of the moment of inertia of 
the spokes is included in that of the counterbalance and 



MOMENT OF INERTIA 109 

boss. The moment of inertia of the wheel with respect 
to its axis of rotation will first be found. The moment 
of inertia consists of: I^ for tire = 415, I^ for rim = l-JU, 
/q for spokes = 246, I^ for hub = T, I^ for counterbalance 
= 344, and for boss = 73. The total moment is 1224. 

With respect to a gravity axis OZ, Fig. 86, in the plane 
of the wheel when the counterbalance is in a position 
where the line joining its center of gravity to the center is 
perpendicular to OZ, we get for the moment of the vari- 
ous parts: loz for tire = 207, loz for rim = 69, loz for spokes 
= 123, loz for hub = 3, loz for counterbalance = 279, and 
for boss = 73. The total moment is 755. In computing 
the moment of the spokes in this case, it was necessarj^ 
to consider that it differs for each spoke. The value 48 
was obtained by computing the moment of inertia of a 
spoke perpendicular to OZ^ multiplying by 20, deducting 
10% for the part of spokes in counterbalance and boss, 
and then dividing the remainder by 2. This, of course, 
is only a reasonable approximation. 

Problem 85. Compute the moment of inertia of a pair of drivers 
and their axle with respect to their axis of rotation. Use the data 
given above and assume the axle as cylindrical, the diameter being 
9^' and the length 68'^ Ans, 2451. 

Problem 86. Compute the moment of inertia of the pair of 
drivers and their axle, given in the preceding problem, with respect 
to an axis midway between the wheels and perpendicular to the axle. 
Consider the counterbalance of both wheels in such a position as to 
give a maximum moment of inertia and the distance between the 
centers of the wheels 60". Ans. 314."). 

Problem 87. Find the moment of inertia of two cast-iron car 
wheels and their connecting steel axle with respect to (a) their axis 
of roti^tion, (b) an axis midway between the wheels and perpendicular 



110 APPLIED MECHANICS FOB ENGINEERS 

to the axle. Consider the car wheels as composed of an outside tread, 
a circular web, and a hub ; each part may be considered a hollow cyl- 
inder with the following dimensions : tread, outside radius 16'^ inside 
radius 14", width o^" ; web, outside radius 14'', inside radius 5^", 
thickness 1.5" ; hub, outside radius 5 J", inside radius 2|", width 8" ; 
axle (considered cylindrical), 5" diameter and 7' 3" long. Distance 
between centers of wheels 60". According to the assumption made 
above, the flange has been neglected, the web is considered a hollow 
disk, and the axle of uniform diameter throughout its length. The 
results will be approximately as follows : (a) 40, (b) 320. 

Problem 88. The value 755 is the greatest value for the moment 
of inertia of a drive wheel with respect to a gravity axis in its plane. 
The least value will be with respect to an axis at right angles to this 
through the centers of gravity of the counterbalance and wheel. The 
student should compute this least moment of inertia. 

Problem 89. In Problem 86 the drivers have been considered as 
having their cranks in the same plane. In practice they are 90° apart. 
Find the moment of inertia with respect to the axis stated when the 
wheels are so placed. 



CHAPTER VIII 



FLEXIBLE CORDS 

66. Introduction. — A cord under tension due to any 
load may be considered as a rigid body. In the analysis 
of problems in which 
such cords are consid- 
ered, the method of 
cutting or section may 
be used. Since the cord 
is flexible (requiring no 
force to bend it), it is 
easy to see that, no 
matter what forces are 
acting upon it, it must 
have at any point the 
direction of the result- 
ant force at that point, 
and so must be under 
simple tension. If the ^'^- ^^ 

cord is curved, as is the case where it is wrapped around 
a pulley, the resultant force is in the direction of the 
tangent. 

Consider, as the simplest case, a weight TF suspended by 
a cord, as shown in Fig. 88 (a). The forces acting on W 
are shown in (6) of the same figure. The cord has been 
considered cut and the force T^ acting vertically upward, 

111 




112 



APPLIED MECHANICS FOE ENGINEERS 



has been used to represent the tension. Summation of 
vertical forces = 0, gives T = TF". When the weight W^ 
is supported by two cords as in Fig. 88 (c), the cords A 
and B are under tension and may be cut. The system of 
forces acting on the point is shown in (c?), where T^ and 
T^ represent the tensions in the cords A and 5, respec- 
tively. ^x=0 and Sy = give 



and 



2^ cos a =2^^^ cos /9 



These two equations are sufficient to determine the 

unknown tensions F and T^^. 

If two weights TTj and W^ are attached to the cord, as 

shown in the case of 
the cord ^5 (7i) (Fig. 
89), each portion is 
under tension. Con- 
sider the cord cut at 
A and D and repre- 
sent the tensions by 




Fig. 89 

T^ and T^ respectively, 



From ^x=0 and 2^ = we have 



an( 



T^ cos 7 = ^2 cos a 

T^ sin 7+^2 sin a^W^+ W^, 



A consideration of the forces .acting at 5, if we call the 
tension in the portion BC^ T^^ gives, when the summation 
of the X and y forces are each put equal to zero. 



and 



y^cos 7= T3COS/3 
2\sin7- 2^3sinyS= W^ 



FLEXIBLE CORDS 



113. 



In a similar way, consider the forces acting on the point 
c, and we have 

T^ cos /3 = 2^2 ^^^ ^ 
and T^ sin ^ + T^ sin a = W<^, 

Of the six equations given above only four are independ- 
ent; consequently, of the six quantities T^^ T^, T^, a, /3, 
and 7, two must be known in order to determine the other 
four. 

In general, if there are n knots such as B and O of 
Fig. 89, with the weights TF^ IFg, TPg, TF^, etc., attached, 
it will be possible to get n + 2 independ- 
ent equations. These will be sufficient 
to determine the tension in each portion 
of the cord and its direction, provided the 
tension at A, say, and its direction are 
known. If the weights are close together, 
the curve takes more nearly the form of a 
smooth curve. Two special cases of this 
kind are discussed in this chapter in Art. 
68 and Art. 69. 

67. Cords and Pulleys. — When a cord 
passes over a pulley, without friction, 
the tension is transmitted along its length 
undiminished. A weight W attached to 
a cord which passes vertically over a pulley 
is raised by a direct downward pull P on 
the other end of the rope. If there is no friction, P is 
equal to TF. In the case of a system of pulleys, as shown 
in Fig. 90, the cord may be considered as under the same 




Fig. 90 



114 



APPLIED MECHANICS FOB ENGINEEBS 



tension throughout and parallel to itself in passing from 
one sheave to the other. It is then possible to cut across 
the cords, just as was done in the case of the bridge truss, 
Problem 47, where the stress was along the member in each 
case. Cutting all the cords at and considering all the 
forces acting on the sheave 5, we get, calling the tension 
in the cord P, 6 P = TF 

or the tension in the cord is TF/6. A consideration of the 
upper sheave gives y= 7P = 7/6 (TT). The various 
cases of cords and pulleys that come up in engineering 
work may be taken up in a similar way, but in any case 
of cutting cords, it must be remembered that all cords 
attaching one part to another must be cut and the tension 
acting along the cords inserted before the principles of 
equilibrium can be applied. The consideration of the 
friction between cords and pulleys will be taken up in 
Chapter XIV. 

68. Cord with Uniform Load Horizontally. — When a cord 
is suspended from two points A and P, Fig. 91 (a), and 
A. [B loaded with a 

uniform load 
horizontally in 
such a way that 
the points of 
attachment of 
the load to the 
cable are very 
close together, 
the cable takes 
Fig. 91 the form of a 




[a) 



T 



ib) 



;rrn J wwij 



FLEXIBLE CORDS 115 

continuous curve. The resultant tension in the cable in 
this case is in the direction of the cable at any point. 
Suppose the cable cut at the points and (7, where is 
the middle point and Q any point between and jB, and 
consider the forces acting on the cut portion. At Q there 
is a tension T making an angle a with the horizontal, Fig. 
91 (6). At 0, the lowest point on the curve, the tension 
(P) is horizontal. The curve is supposed loaded with a 
uniform load of W pounds per linear foot, so that Wx 
represents the total horizontal loading. Writing down the 
equations Sa: = and 2y = 0, we obtain 

P= ^cosa, 

Wx = T sin a, 

Wx 

and by division tan a = — — - . 

But tan a = ^, giving ^ = ^. 
ax dx P 

Therefore, y = or x^ =4^^' 

which is the equation of the curve taken by the cable under 
the assumed loading. This is a parabola. The deflection 
of the curve at any point can be found by putting in the 
value of X for that point and solving for y. If I equals 
length of span and d the maximum deflection at the center, 

then d = — — . 
8P 

The length of the cable may be found by considering 

the formula ^^ 



116 APPLIED MECHANICS FOR ENGINEERS 

where ds is measured along the curve. The length OB^ 
or semi-length of the cable for a span Z, is, since 

ds^ = dx^ + (7^2, 

op 

From the equation of the curve x^ = v. we have 

dji_ _ Wx 

dx'~'~P' 

Therefore, 






1^2 + ^ 



Expanding the two terms of the above expression into 
infinite series and adding like terms, we may express the 
total length of the parabola in terms of Z, TF, and P : 

Total length - 1 + ^^ - ^^-\ + • • 
24 P*^ 708 P* 



or in terms of I and c? 



Total length = I -f ^^^ - '^^^ + 

70 



8^ 32^* 
3^ ht 



In general, the convergence of either of the above series 
will be sufficiently rapid that only the first two terms 
need be used. In such cases they will be found con- 
venient for computation. 



FLEXIBLE CORDS 



117 



The point of maximum tension in the cable is determined 
by considering the equation P = T cos a ov T = 



cos a 

It is easy to see that T is greater, the smaller the value 
of cos a, that is, the larger the value of «, and this is 
greatest at the points of attachment A and B, Tliis is 
the problem of the suspension bridge where tlie weight of 
the cable is neglected. 

69. Equilibrium of a Flexible Cord Due to its Own Weight. — 
Where the loading along a flexible cord or cable is uni- 
form, as in the case where the cable bends due to its own 
weight, the shape of the 
curve taken is no longer 
a parabola, as will be 
shown in what follows. 
Suppose a portion of the 
cable, 00, Fig. 92, with 
its load, cut free, and let 
the tensions in and C ( is the middle point and O is any 
point between and jB) be P and T respectively. Then, 
2x = and 2^/ = give P = T cos a and W8= T sin a, 
and from these two equations, by division, we have 




tan a = 



Ws 



dy 
dx 



or -^ = 



Ws 
~P' 



where 8 represents distance along the curve, and is re- 
lated to X and y in such a way that ds'^ = do^ + dy'^. 
Eliminating dy between this and the previous equation, 

we obtain 

ds P ds 

W ' 



dx = 



>|i + 



p2 



4 



p2 

jy2 



+ s 



,2 



118 



APPLIED MECHANICS FOR ENGINEERS 



Therefore, ^ = w" ^^Se ( ^ + \ "ttT^ + "^ 



=w''^^ 



s + 



4 



F2 



+ s^ 



W 



(«) 



This gives a relation between x and s. A relation be- 
tween y and s may be obtained by eliminating dx between 

the equations -^ = -— - and d8^ = dx^ + dy^. This gives 
dx P 



dy=^ 



sds 






+ 



Therefore, 



y 



=V^+4=V- 



h S^ — ■ 

W^ w 



(5) 



Eliminating s between (a) and (6), we get the equation 
of the curve taken by the cable to be 



4- — = — 



(WX VVX\ 



(0 



This is the equation of the catenary with the origin at the 
vertex and the y-axis the axis of symmetry. 

If the length of span is Z, the maximum deflection of the 
cable at the center may be determined by substituting 

2: = - in (c). The semi-length of the cable may be found 

I 
from (a) by substituting x = ~ and solving for s. For 

this purpose (a) may be written in the exponential form ; 



FLEXIBLE CORDS 119 

remembering that a"^ = n may be written m = log^ n, we 
then have 

p 

Since P = T cos a or T= , it is evident that the maxi- 

COS a 

mum tension occurs at the supports A and B. 

70. Representations by Means of Hyperbolic Functions. — 
Equations (c) and (cZ) may be expressed in a simpler form 
by using the hyperbolic functions, remembering that the 
hyperbolic sine and cosine are expressed 



. . Wx 
sinh — 

, Wx 
cosh — 

P 2 

We have for equation ((?) 

, P P , Wx 

'"-w^w'^'^'-p-^ 

and for equation (c?) p . ^ W^ 

'"If 7^- 

It is evident that for rapidity of computation of s and 
^, tables giving the values of the hyperbolic sine and 

cosine, for various values of — ^, would be convenient. 

For this reason a table is given in Appendix I, and the 
student is requested to use this table in solution of the 
problems. 



e 


Wx 

p 


— e 


Wx 
P 






2 


9 


e 


Wx 

p 


+ e~ 


Wx 

P , 



120 



APPLIED MECHANICS FOR ENGINEERS 



Problem 90. A suspension bridge as shown in Fig. 93 has a span 
of 1200 ft. and the cable a maximum deflection at the center d = 120 
ft. The weight of the floor is 2 tons per linear foot. Find the 




Fig. 93 

equation of the cable and the tension at and at B. If the safe 
strength of cable is 75,000 lb. per square inch, find the area of wire 
section of cable necessary to support the floor. 

Problem 91. Find the length of the cable in the preceding 
problem. 

Problem 92. A flexible wire weighing ^ lb. per foot is sup- 
ported by two posts 200 ft. apart. The horizontal pull on the 
wire is 500 lb. Find the deflection at the center and the length of 
the wire. 

Problem 93. What pull will be necessary in Problem 92 so that 
the greatest deflection will not be greater than 6 in.? What is the 
length of the wire for this case? 

Problem 94. Find the tension in the wire of Problem 92 at the 
supports. 

In practice use is often made of the fact that the exponential func- 
tion may be expanded into an infinite series, so that 

P P .Wx 

V H = — cosh ■ 

^ W W P 



FLEXIBLE CORDS 121 

Wx 



may be written, remembering the meaning of cosh 



■). 



^ 2r2S^r3^ 



In a similar way we may write 

P ( 2 Wx WV W^x^ 

2W\ P 3P3 3.4.5P6 "^ 

W2x3 

Wl 
Since tan a at the supports may be written tan a = sinh , we 

may write it as the following series : 



ir 48 i^3 2 3 4 5 321^5 

When the series are rapidly convergent, only the first terms need 
be used, so that 

TF.r2 , W^x^ 

s = x -\ , 

2.3P2' 



and tana = I^+^^-^ 



P 6P3 



When y = c?, x = -, so that 



d —- —— + 



8P 2.3.4.I6P8 

or approximately _ JVli . or, 7^ - Wl'^ 

d-^-^,andsoJ>_-^. 



122 APPLIED MECHANICS FOR ENGINEERS 

Also at the supports tan a is approximately 

jn 4 €i 



When X — -t 



^"^^ = 21> I 



s — - A = - H . 

2 48P^ 2 6Z 



The total length of the cable or wire may then be expressed as 



Total length = Z + 



3« 



The student should make use of these formulae in solving the pre- 
ceding problems. 



CHAPTER IX 

MOTION IN A STRAIGHT LINE (RECTILINEAR MOTION) 

71. Velocity. — The velocity of a body is its rate of 
motion. If the velocity is constant (uniform), it may be 
defined as the ratio of the distance passed over to the 
time spent in passing over that distance. If the velocity 
is variable, the velocity at any instant is the velocity that 
the body would have if at that instant the motion should 
become uniform. Speed is sometimes used instead of 
velocity, especially in speaking of the motion of machines 
or parts of machines. Speed, however, involves only the 
rate of motion without reference to the direction of 
motion, while velocity involves both rate of motion and 
the direction in which the motion takes place. Since con- 
stant velocity is the ratio of distance to time, it may be 
represented as 

8 

t 

The units for measuring velocity are those of distance 
and time, usually feet and seconds. Thus a body has a 
velocity of k feet per p 

second or a train has ^ 9 7 

a velocity of k miles F^ i' -A*— H 

1 Fig. ^ 

per hour. 

A formula for expressing the relation between velocity, 
distance, and time for variable velocity may be derived by 

123 



124 APPLIED MECHANICS FOR ENGINEERS 

referring to Fig. 94. Suppose a body to have moved 
from to P over the distance s with variable velocity. 
Let V be its velocity at P and t the time. In moving to 
another position P^ distant As, the velocity changes by an 
amount Av and the time by an amount Af, so that at P^ 

V + Av = — ; 

A^ 

as Av^ As, and At approach zero as a limit, 

that is, the velocity is the first derivative of the distance with 
respect to time. 

72. Acceleration. — Acceleration may be defined as the 
rate of change of velocity. If the velocity changes by 
equal amounts in equal times, tlie acceleration is said to 
be constant or uniform^ otherwise it is variable. Constant 
acceleration, then, is the ratio of the velocity to time; rep- 
resenting this acceleration by a^, we have 

The units used are those of velocity and time, and since 
velocity is usually expressed in terms of feet and seconds 
or feet per second, acceleration is usually expressed in 
terms of feet per second per second. This is sometimes 
expressed as feet per square second or simply as feet per 
second, it being understood that the time must enter 
twice. 

Since the acceleration equals the rate of change of 
velocity at any instant, we may write 



MOTION IN A STRAIGHT LINE 125 

^ dv d^s 

d = — = ? 

dt dt2 

where a represents the variable acceleration. Since 

v = — and a = — , vdv = ads, by eliminatine^ dt. 
dt dt J & 

( ^- ^^-- 

73. Constant Acceleration. — When the acceleration is 
constant, we have the relation dv = a^dt^ a^ representing 
the constant value of a, and therefore 






. 




%/ Vq c/0 






V = a^« + Vo ; 


and since 
or 




ds 

V = —i 

dt 
£ds = ajjtdt + 


In a similar 


way 


the relation 
vdv = a^ds 


gives 




£vdv = a,j^ds; 


therefore 




v^ vl 

2 2 ~^«*' 


or 







These equations of motion give the velocity in terms of 
time, the distance in terms of time, and the distance in 
terms of velocity. 



126 APPLIED MECHANICS FOR ENGINEERS 

74. Freely Falling Bodies. — Bodies falling toward the 
earth near its surface have a constant acceleration. It is 
usually represented by ^ and equals approximately 32.2 ft. 
per second squared. The value of g varies slightly with 
the height above the sea level and the latitude, but for 
the purposes of engineering it may usually be taken as 
32.2. The equations of motion for such bodies are, then, 

v=gt + VQ, 
8 = ^gt^ + VQt, 

V^ — Vq 

If the body falls from rest, v^ = 0, and the equations of 

motion become 

v=gt, 

This latter is often written v^ = 2gh^ where h= s. 

75. Body Projected vertically Upward. — When a body 
is projected vertically upward from the earth, the accelera- 
tion is constant and equals —g. If the velocity of pro- 
jection is ^Q, the equations of motion are 

V = — gt + v^ , 

2 2 

Problem 95. A body is projected vertically downward with an 
initial velocity of 30 ft. per second from a height of 100 ft. Find 
the time of descent and the velocity with which it strikes the ground. 



MOTION IN A STRAIGHT LINE 127 

Problem 96. A body falls from rest and reaches the ground in 
6 sec. From what height does it fall, and with what velocity does 
it strike the ground ? 

Problem 97. A body is projected vertically upw\ard and rises to 
the height of 200 ft. Find the velocity of projection v^ and the 
time of ascent. Also find the time of descent and the velocity with 
which the body strikes the ground. 

Problem 98. A stone is dropped into a well, and after 2 sec. 
the sound of the splash is heard. Find the distance to the surface 
of the water, the velocity of sound being 1127 ft. per second. 

Problem 99. A man descending in an elevator whose velocity 
is 10 ft. per second drops a ball from a height above the elevator 
floor of 6 ft. How far will the elevator descend before the ball 
strikes the floor of the elevator ? 

Problem 100. In the preceding problem, suppose the elevator 
going up wdth the same velocity, find the distance the elevator goes 
before the ball strikes the floor of the elevator. 

76. Newton's Laws of Motion. — Three fundamental laws 
may be laid down which embody all the principles in 
accordance with which motion takes place. These are 
the result of observation and experiment and are known 
as Newton's Laws of Motion, 

First Law, Every body remains in a state of rest or 
of uniform motion in a straight line unless acted upon by 
some unbalanced force. 

Second Law. When a body is acted upon by an unbal- 
anced force, motion takes place along the line of action 
of the force, and the acceleration is proportional to the 
force applied. 

Third Law. To every action of a force there is always 
an equal and opposite reaction. 



128 APPLIED MECHANICS FOR ENGINEERS 

The first law has already been made use of, and also 
the third law — see articles of Chapter II. 

The second law states that in case the system of forces 
acting on the body is unbalanced, the motion is accelerated. 
Motion takes place in the direction of the resultant force 
with an acceleration proportional to the force. It also 
implies that each force of the system produces or tends 
to produce an acceleration in its own direction propor- 
tional to the force. That is to say, each force produces 
its own effect, regardless of the action of the other forces. 

As a result of this latter fact, if a body is acted upon 
by a force P and the earth's attraction Gr^ we have 

P: a = a:g, 

where Gr is the weight of the body, g the acceleration of 
gravity, and a the acceleration due to the force P. From 
this it follows that 

g 

that is, the accelerating force equals the mass (see Art. 7) 
times the acceleration. 

77. Motion on an Inclined Plane. — A body (see 

Fig. 95), of weight (r, moves 
down an inclined plane, with- 
out friction under the action 
of a force Gi sin 6. The 
acceleration down the plane 
equals the accelerating force 
Fig. 95 divided by the mass (see Art. 

76) = j^ — = g sin 6, The acceleration is constant. 

1 




MOTION IN A STRAIGHT LINE 129 

The equations of motion for such a case, then, are (see 

Art. 73) 

v = {g sin 6) f + v^^ 

8 = \g (sin 0) f3 + v^t, 

s = ?. 

2 gr sm 6 

If the body starts from rest down the plane, Vq = 0. If 
it be projected up the plane with an initial velocity v^, the 
acceleration equals —g sin 6. 

Problem 101. A body is projected up an inclined plane which 
makes an angle of 60° with the horizontal with an initial velocity of 
12 ft. per second. Neglecting friction of the plane, how far up the 
plane will the body go? Find the time of going up and of coming 
down. 

Problem 102. A body is projected down the plane given in the 
preceding problem with a velocity of 20 ft. per second. How far 
will it go during the third second? 

Problem 103. Suppose the body in the preceding problem 
meets a constant force of friction F = 10 lb. What will be the 
acceleration down the plane? How far will it go during the second 
second ? 

Problem 104. A boy who has coasted down hill on a sled has a 
velocity of 10 mi. per hour when he reaches the foot of the hill. 
He now goes on a horizontal, meeting a constant resistance of 25 
lb. If the combined weight of the boy and sled is 75 lb., how far will 
he go before coming to rest ? 

Problem 105. Suppose that in the preceding problem the boy 
weighs 65 lb. and the sled 10 lb., and that the boy can exert a force 
of 20 lb. horizontally to keep him on the sled. Will the boy 
remain on the sled when the latter stops, or will he be thrown 
forward ? 



130 



APPLIED MECHANICS FOR ENGINEERS 




G^ = 20 LBS. 



Fig. 96 



Problem 106. A body whose 
weight G = D lb. is being drawn 
up an inclined plane as shown in 
Fig. 96 by the action of the 
weight G = 20 lb. Suppose the 
resistance offered by the plane 
F = 10 lb., and that G starts 
from rest. How far up the 
plane will G go in 6 sec? 



Problem 107. Two weights, Gi = 6 lb. and G2 

attached to an inextensible cord 
w^hich runs over a pulley, are acted 
upon by gravity ; no friction ; mo- 
tion takes place. Find the tension 
in the cord, and the acceleration. 
Consider G2 and Gi separately 
with the forces acting upon them, 
and call the tension in the cord T. 
Then apply the principle " acceler- 
ating force equals mass times acceler- 
ation, '^ 



10 lb., Fig. 97, 




T 



G. 



Problem 108. 




Fig. 98 

ing problems, 
descending? 



An elevator. Fig. 98, whose weight is 2000 lb. is 
descending with a velocity, at one instant, of 2 
ft. per second, and at the next second it has a 
velocity of 18.1 ft. per second. Find the tension 
T in the cable that supports the elevator. 

Problem 109. Suppose the elevator in preced- 
ing problem going up with the same acceleration. 
Find the tension in the cable if the elevator starts 
from rest and attains its acceleration in 3 sec. 

Problem 110. A man can just lift 200 lb. 
when standing on the ground. How much could 
he lift when in the moving elevator of the preced- 
(a) when the elevator was ascending? (b) when 



MOTION IN A STRAIGHT LINE 



131 



Problem 111. Two weights, G and G\ are connected by an 
inextensible flexible cord 
that passes over a friction- 
less pulley, as shown in 
Fig. 99. G = 20 lb., G' = 
100 lb., and there is no 
friction on the plane. Find 
the tension in the cord and 
the acceleration of the two 
bodies. 




Fig. m 



Problem 112. A 30-ton car is moving with a velocity of 30 mi. 
per hour on a level track. The brakes refuse to work. Ho^y far 
will the car go after the power is turned off before coming to rest, 
if the friction is .01 of the weight of the car? 

78. Variable Acceleration. — It has already been shown 

fl Q Cm 1) Cr^ 

that V = — . a = — =— ^, and vdv = ads. These relations 
dt dt dt^ 

hold true no matter whether the acceleration is constant 
or variable. If the acceleration is constant^ the equations 
of motion are those that have already been worked out 
(see Art. 73), and by simple substitution in these equa- 
tions it is possible to find the velocity in terms of the 
time, the distance in terms of time, and the distance in 
terms of velocity. 

If the acceleration is variable, it is necessary to work 
out the equations of motion for each case. This may be 
done, when it is known how a varies, by means of either 
of the equations. 



a = 



or 



d^ 
dfi' 
vdv = ads. 



The latter equation will usually give the beginner less 
difficulty. 



132 APPLIED MECHANICS FOR ENGINEERS 

79. Harmonic Motion. — Let it be supposed that a body 
is moved by an attractive force which varies as the dis- 
tance. That is, the attractive force is proportional to 
the distance. Then the acceleration is also proportional 
to the distance. 

Let the acceleration = — ks. 

Then vdv = — ksds, 

and I vdv = —Jc \ sds ; 

therefore v^ — v^^ — ks\ 

where Vq is the initial velocity when s equals zero and 
k is the factor of proportionality, determinable in any 
special case. This equation gives the relation between 

the velocity and distance. Since v = Vv^ — ks^^ it is 

evident that v = when V^ - s= v^. This means that the 
body comes to rest when s has reached a certain value, 

viz. — ^ • From the original assumption, a= — ks^ it is 
■y/k 

seen that the acceleration is greatest when s is greatest, 

Vk 
when s= 0. 

To get the relation between distance and time, the 

equation v = Vt^g — ks^ may be put in the form 

ds 



V » 

that is, when s = — % ; and is least when s is least, that is. 



Vv2 __ J^g2 



= dt, 





from which, — =. sin"^ '— = f, 

■Vk ^0 

or -4. sin -y/kt = s. 

^k 



MOTION IN A STRAIGHT LINE 133 

This relation between the distance and time shows that 

as t increases s chang^es in value from — ^_ to ^, as- 

^k Vk 

suming all values between these limits, but never exceed- 
ing them, since sin ^kt can never be greater than + 1 or 
less than — 1. The motion is, therefore, vibratory or 
periodic, and is known as harmonic motion. The complete 

period m tins case is — z. • 

Vyfc 

The relation between velocity and time may be found 
for this case by differentiating the last equation with 
respect to time. Then, 

V = Vq cos Vkt 

This shows that v^ is the greatest value of v. 

This motion is usually illustrated by imagining a ball 
attached by means of two rubber bands or springs, since 
the force exerted by either ^ ^ 

of these is proportional to L ^ _J 

the elongation, to two pins, IS 

as shown in Fig. 100. As- ^'^- ^^^ 

suming that there is no friction and that the ball is dis- 
placed to a position B by stretching one of the rubber 
bands, when released it continues to move backward and 
forward with harmonic motion. 

Problem 113. Suppose the ball in Fig. 100, held by two helical 
springs, to have a weight of 10 lb. and that it is displaced 1 in. 
from O. The two springs are free from load when the body is at O. 
The springs are just alike, and each requires a force of 10 lb. to 
compress or elongate it 1 in. Find the time of vibration of the 
body and its velocity and position after \ sec. from the time when 
it is released. It has been found by experiment that the force neces- 



134 



APPLIED MECHANICS FOR ENGINEERS 



sary to compress or elongate a helical spring is proportional to the 
compression or elongation. 

80. Motion with Repulsive Force Acting. — Suppose the 
force to be one of repulsion and to vary as the distance ; 
then a = ks^ and vdv = ksds^ so that 



s = -\(e^~''^ - e-^''A = -^ sinh ^/kf 
2Vk\ J Vyfc 

These equations show that as t increases s also increases 
and the body moves farther and farther away from the 
center of force. The motion is not oscillatory. 

81. Motion where Resistance varies as Distance, — If a 

©body w^hose weight is 644 lb. falls freely 
from rest through 60 ft. and strikes a 
-. . resisting medium (a shaft where friction 

on the sides equals 2 ^=10 times the 
distance; see Fig. 101), since accelerating 
force equals mass times acceleration, 




a-2F a-ios 



a = 



M 



a 

9 



= 9- 



2* 



It is required to find (a) the distance 
the body goes down the shaft before 
coming to rest ; (J) the distance at which 
the velocity is a maximum; ((?) the total 
time of fall ; (c?) the velocity at a distance 
of 10 ft. down the shaft. After striking 

the shaft the relation between velocity and distance is 

as follows : 



Fig. 101 



MOTION IN A STRAIGHT LINE 



135 



Jvdv = I 



9 



ds. 



The remainder of the problem is left as an exercise for the 
student. 



Problem 114. A ball whose weight is 32.2 lb. falls freely from 
rest through a distance of 10 ft. 
and strikes a 400-lb. spring, Fig. 
102. Find the compression in 
the spring. It is to be under- 
stood that a 400-lb. spring is such 
a spring that 400 lb. resting upon 
it compresses it one inch, and 
4800 lb. resting on it compresses 
it one foot, if such compression 
is possible. After the ball strikes 
the spring it is acted upon by the 
attraction of the earth and the 
resistance of the spring. The ac- 

O _ 4800 5 



celeration a is then 



M 



where s is measured in feet. The 
relation between velocity and dis- 
tance is then obtained from the 
relation, 






4800 s) ds. 




Fig. 102 



Problem 115. A 20-ton freight car, Fig. 103, moving with a 
velocity of 4 mi. per hour strikes a bumping post. The 60,000-lb. 
spring of the draft rigging of the car is compressed. Find the com- 
pression s. Assume that the bumping post absorbs none of the shock. 

Problem 116. Suppose the car in the preceding problem to be 
moving with a velocity of 4 mi. per hour, what should be the 



136 



APPLIED MECHANICS FOR ENGINEERS 



strength of the spring in the draft rigging so that the compression 
cannot exceed 2 in. ? 

Problem 117. After the spring in Problem 114 has been com- 
pressed so that the ball comes to rest, it begins to regain its original 



4 Ml. PER HR. 




3wwww\/\D 





Fig. 103 



form. Find the time required to do this and the velocity with which 
the ball is thrown from the spring. 



82. Motion when Attractive Force varies inversely as 
Square Distance. — This is the case 

I of motion, Fig. 104, when two bodies 

G in space are considered, since in such 

cases the attractive force varies di- 
rectly as the product of their masses 
and inversely as the square of the dis- 
tance between them. The same at- 
traction holds between two opposite 
poles of magnets or between two 
bodies charged oppositely with elec- 
tricity. 

— k 
Suppose the acceleration = — -- and that the velocity is 




Fig. 104 



zero. 



MOTION IN A STRAIGHT LINE 137 

vdv= — I .,^7s, 



so that 
and 









VSqS — S^ 



-^vers 1 ^ 

9 ^ -> 



The time required to reach the center of attraction O 
from the position of rest is obtained by putting s = 0. 

This gives ^ = —f^\2. 

It is seen that when s = the velocity is infinite, and 
therefore the body approaches the center of attraction 
with increasing velocity and passes through the center, to 
be retarded on the other side until it reaches a distance 
— «Q. The motion will be oscillatory. 

If one of the bodies is the earth, of radius r, and the 
other is a body of weight Gr falling toward it, the equa- 
tions just derived hold true. In this case it is possible 
to determine k. The attraction on the body at the sur- 
face of the earth is (7, and at a distance s is F^ so that 

F= (xf — ). The acceleration is therefore = — J — 

This gives k, then, equal to r^g. 

Substituting these values in the above equation, we find 



.=# 



gr^ 'VsnS — s^ 



'SI 



h « 



When s = r vt the earth's surface, 



138 APPLIED MECHANICS FOR ENGINEERS 



If Sq= cc, v= V2 gr. 

But this is a value of v that cannot be obtained, since 

Sq cannot be infinite. So that the velocity is always less 

than V2 gr. It is interesting to notice here that if a 

body were projected from the earth with a velocity 

greater than V2 gr^ it w^ould never return, provided there 

were no atmospheric resistance. Substituting ^= 32.2 

and r = 3963 mi., 

V = 6.7 mi. per second. 

This is the greatest velocity that a body could possibly 
acquire in falling to the earth, and a body projected 
upward with a greater velocity would never return 
(neglecting resistance). 

If the body falls to the earth from a height A, the veloc- 
ity acquired may be obtained from the foregoing by put- 
ting s = r and SQ = h + r ; then 



. 2 grh 
^r + h 

If h is small compared to r, this may be written, without 

serious error, /- — - 

V = yzgh, 

which is the formula derived for a freely falling body in 
Art. 74. 

83. Motion of a Body through the Atmosphere. — When a 
body such as a raindrop moves through the air, the resist- 
ance varies approximately as the square of the velocity. 
Suppose a body of weight Gr projected vertically upward 
in such a medium and let the resistance be Ii = Ma = 



a=-g[l-:^v^ 



e-- 



-/'.. 



H U. >• 7 



MOTION IN A STRAIGHT LINE 



139 



And the relation between velocity and distance is ex- 
pressed by the equation 

vdv= — g\l + -^vAds, 



or 



^ a 



a 1 

2k ° 



1 + 


^ 21 

(7^ 


1 + 





This gives the relation between velocity and distance. It 
is left as a problem for the student to determine the 
relation between distance and time for this case. Find 
the greatest height to which a body will rise and the 
velocity with which it strikes the ground upon re- 
turning. Compare this velocity with the velocity of 
projection. 

84. Relative Velocity. — When we speak of the velocity 
of a body, it is understood that we mean the velocity of 
the body relative to the earth, more particularly the point 
on the earth from which the motion is observed. Since 
the earth is in motion, it is evident that velocity as gen- 
erally spoken of is not absolute velocity, and since there 
is nothing in the universe that is at rest, all velocities 
must be relative. In everyday life, however, we think of 
velocities referred to any point on the surface of the earth 
as being absolute. 

A person walking on the deck of a boat, for example, 
has a velocity relative to the earth and a velocity relative 



140 



APPLIED MECHANICS FOR ENGINEERS 



to the boat ; the former is usually spoken of as the absolute 
velocity, and the latter the relative velocity. Or, suppose 
the case of a man standing on the deck of a boat moving 
south with a velocity v while the wind blows from the 
east with a velocity Vy It is required to find the velocity 
of the wind with respect to the man, or, in other words, 
the apparent direction and velocity of the Avind as ob- 
served by the man. Referring to Fig. 105, 
we represent the velocity (see Art. 85) of 
the boat with respect to the earth by v^ 
and the velocity of the wind with respect 
to the earth by i^j, then V represents the 
velocity of the wind with respect to the man ; 
that is, the wind appears to the man to be 
coming from the southeast. The velocity 
V was obtained by reversing the arrow rep- 
resenting the velocity of the boat and find- 
ing the resultant of this reversed velocity 
and the velocity v-^ of the wind. 
If v-^^ be considered as the velocity with respect to the 
earth of a man walking- across the deck of a steamer mov- 
ing with a velocity v^ then F' represents the velocity of the 
man with respect to the boat. 




Fig. 105 



Problem 118. An ice boat is moving due north at a speed of 
60 mi. per hour, the wind blows from the southwest with a 
velocity of 20 mi. per hour. What is the apparent direction and 
velocity of the wind as observed by a man on the boat ? 



Problem 119. A man walks in the rain with a velocity of 4 
mi. per hour. The rain drops have a velocity of 20 ft. per second 
in a direction making 60° with the horizontal. How much must the 
man incline his umbrella from the vertical in order to keep off the 



MOTION IN A STRAIGHT LINE 141 

rain : (a) when going against the rain, (J>) when going away from 
the rain V If he doubles his speed, what change is necessary in the 
inclination of his umbrella in (</) and (A) ? 

Problem 120. The light from a star enters a telescope inclined 
at an angle of 45° with the surface of the earth. The velocity of light 
is 186,000 mi. per second and the earth (radius 4000 mi.) makes 
one revolution in 24 hr. What is the actual direction of the star 
with respect to the earth? This displacement of light due to the 
velocity of the earth and the velocity of light is known as aberration 
of light. 

Problem 121. A man attempts to swim across a river, 1 mi. 
wide, which is flowing at the rate of 4 mi. per hour. If he can swim 
at the rate of 3 mi. per hour, what direction must he take in swim- 
ming in order to reach a point directly across on the opposite shore ? 

Problem 122. A train is moving with a speed of 60 mi. per 
hour, another train on a parallel track is going in the opposite direc- 
tion with a speed of 40 mi. per hour. What is the velocity of the 
second train as observed by a passenger on the first ? 

Problem 123. A man in an automobile going at a speed of 40 
mi. per hour is struck by a stone thrown by a boy. The stone has 
a velocity of 30 ft. per second and moves in a direction perpendicu- 
lar to the direction of motion of the automobile. With what velocity 
does the stone strike the man ? 

Problem 124. A locomotive is moving with a velocity of 40 
mi. per hour. Its drive wheels are 80 in. in diameter. What is the 
tangential velocity of the upper point of the wheels with respect to 
the frame of the locomotive ? What is the tangential velocity of the 
lowest point? 



CHAPTER X 



CURVILINEAR MOTION 



85. Representation of Velocity and Acceleration. — It has 

been shown (Art. 71) that velocity is measured in terms 
of feet per second, miles per hour, or in general in terms 
of the units of distance and units of time. The velocity is, 
moreover, in a given direction and may accordingly be 
represented by an arrow just as forces may be so repre- 
sented. It follows then that velocity arrows may be 

added algebraic- 
ally if parallel and 
if such addition is 
not inconsistent 
with the prob- 
lem. They may 
be resolved into 
components or 
combined to form 
resultants (see 
Art. 11 and Art. 
12). In case the body moves in a curve it is often desir- 
able, instead of dealing with the resultant velocity along 
the tangent, to deal with the components of that velocity 
along the two coordinate axes. Thus if v is the velocity 
along the tangent (Fig. 106), 'Vx=v cos a and ^y = v sin a are 

142 




Fig. 106 



CURVILINEAR MOTION 143 

the component velocities along the axes x and y respec- 
tively. In a similar way if we know the velocity 
of a body along the a:-axis, v^^ and the velocity along 
the y-axis, Vy^ we find the resultant velocity to be 
^ = V v^+vl and its direction with the 2:-axis such that 

tan a = ~- 

Accelerations have been seen to be measured in terms 
of units of distance and units of time, in particular in 
terms of feet per (second)^ (see Art. 72). An accelera- 
tion may be represented by an arrow, the length of the 
arrow representing the number of feet per (second)^ and 
the direction of the arrow giving the direction of the 
acceleration. Since arrows represent accelerations, the 
acceleration arrows may be treated just as velocity arrows. 
That is, they may be added algebraically if parallel, 
or added and subtracted geometrically if intersecting. 
Or we may say that the parallelogram law holds for 
accelerations. Referring to Fig. lOG, it is seen that the 
resultant acceleration, a, of the body moving in the curve 
y =zf(x) is directed toward the concave side of the curve 
in the direction of the resultant force. This is evident 
from Newton's Law, which states that the acceleratioyi is 
proportional to the resultant force and in the same direction. 
Let a be the resultant acceleration, then the accelerations 
along the two axes are ax=acos 9 and ay=a siu 8, respec- 
tively. In a similar way if we know the accelerations 
along the two axes, a^ and a^, the resultant acceleration 
a = Va|~+~a2, and its direction is given by the equation 

tan^=^. 
ar 



14-4 APPLIED MECHANICS FOR ENGINEERS 

86. Tangential and Normal Accelerations. — Suppose a 
body (Fig. 106) moves in any curve y =f{x) and 
that at a certain point P it has a resultant velocity v 
and a resultant acceleration a, v acts along the tangent, 
which at this point makes an angle a with the a::-axis and a 
acts along the line of action of the resultant force on the 
concave side of the curve. It is seen from the figure that 
v^=^ V cos a^ Vy=^ V sin a, a^: = a cos 0^ ay = a sin 0^ so that 
V = -y/vl + v^ and a = Va| + a^. 

It is usually convenient in curvilinear motion to con- 
sider the acceleration along the tangent and normal ; that 
is, a^ and a^. Since the tangent and normal are at right 
angles, it is evident that a = V^^ _|_ ^2^ When it is 
remembered that v acts along the tangent, it is evident 

that the tangential acceleration at = — , or, 

dt 



dt '^ ' ^ dt\\jtj ^\dtj 

fcPx dx d^y dy 



^dx\^(dy\^\dt^dt dt^ dt 
\\dt) '^\dt) 



)■ 



Since 



= - (^a^Vj, + ayVy) = a^ cos a + ay sin a^ 

V 

^ == cos a and -^ = sm a. 

V V 



It now remains to find the normal acceleration. It has 
been shown that a = Va^ + a^ ; therefore, a„ = Va^ — a^. 
Substituting" the value of a^ = Ca'^ + a^), and the value of 
a^ just found, we have as the va-lue of the normal accelera- 
tion. 



CUnVILIXEAR MOTION 145 



a,i = ay cos cc— aj. sm a. 



The norinal force and tangential force may now be found 
by multiplying by the mass iff, giving, 

Normal force = 3Iay cos a — Ma^ sin a; 
Tangential force = ilfa^ cos « + ifcTa^ sin a. 

It is usually, however, more convenient to have the 
normal and tangential forces expressed in terms of tlie 

fi 1) (ii) 

velocity. Since a^ = — , the tangential force = M — To 

dt dt 

express the normal acceleration in terms of velocity it is 
necessary to write. 



^71 = 



_ fd?']/ dx d'^x dii 



dfi dt dt^ dtj \fdx\^ fdy 



mr-m 



[ 



cPy dx _ d^x dy 
liflTt dfidt /ds\^ 



dtJ ydtj _ 



= - — = — . (bee note. ) 



Note. Since ?/ is a function of x and both x and y are functions 

of t^ we may write, 

dy dP-y dx __ d^x dy 

dy dt , r/2// dt- dt dt'^ dt 

-f- = —- and —4 = 

dx dx_ dx^ 

dt 



m 



The expression for the radius of curvature of a curve whose equation 
is y =f(x) is 

[(f)'^(g)']' 

^ ^/-// dx d'h: dy 
dt'^ dt dt^ dt 



146 



APPLIED MECHANICS FOR ENGINEERS 



The normal force and tangential force may now be 
written, 

Normal force 



3Iv^ 



Tangential force = M^. 

dt 

For all curves except the circle p the radius of curva- 
ture varies from point to point. In the circle, however, it 
is the radius, and is therefore constant. In this case the 
normal force is usually called the centripetal force. 

87. Uniform Motion in a Circle. — A body moving with 
constant velocity in the circumference of a circle is acted 
upon by one force, the normal or centripetal force, and this 



equals 



Mv' 



That this is true is evident when it is re- 



membered that the tangential velocity is constant, thus 

making the tangential accelera- 
tion zero. An illustration of 
uniform motion in a circle is 
seen in the case of the simple 
governor shown in Fig. 107. 
When the velocitv is constant, 
then a, A, and r are constant. 
Let T be the tension in the rod 
supporting the ball, then, since 
there is no vertical motion 2y = 
0, so that T cos a = Gr, Consid- 




FiG. 107 



ering the normal force, we have T sin a 



Mv' 



so that 



tan a = — . From these equations T may be found for any 
gr 

values of a and r. 



C UR VIL INK A R MO TION 



147 



Problem 125. The weighted governor shown in Fig. 108 is rotated 
at such a speed that a = 30°. Find the forces acting on the longer 
rods and the stress in the 
shorter rods. The connec- 
tions are all pin connec- 
tions. 

Problem 126. A type 
of swing is shown in Fig. 
109. A revolving central 
post supported by wires A 
and B carries six cars G, 
each suspended from cross 
arms D by means of cables 
50 ft. long. When the swing 
is at rest, the cars hang ver- 
tically and a = ; as the 
speed of rotation increases* 
a becomes larger. Sup- 
pose the car and its load of 
four passengers to weigh 
1000 lb., and the speed to 




=» 10 LBS. 



20 LBS. 



Fig. 108 



be such that a = 30°, find the tension in the cables supporting the 
cars. Assume that a single car is carried by one cable. 

Problem 127. The 

same principle that has 
been seen to hold for 
motion in a circle en- 
ables us to solve the 
problem that comes up 
in railroad work. 
When a train goes 
around a curve, it is 
desirable to have the 
outer rail raised suffi- 
ciently so that the 




Fig. 109 
wheel pressure will be normal to the rails. It is really the same 



148 



APPLIED MECHANICS FOR ENGINEERS 




Fig. 110 



problem as Problem 126, where the sustaining cable is replaced by 
a track (see Fig. 110). Let r be the radius of curvature, v the veloc- 
ity of the car, of weight G. Show that the superelevation of the 

outer rail is given by tan a = — , and so 

where d is the distance between the 
rails in feet, r the velocity in ft. per 
second, g is 32.2, r is the radius of 

curvature in feet, and li is the superelevation of the outer rail in feet. 

This height may be expressed, approximately, as follows ; 

3r 

where A and r are in feet and v^ is the velocity in miles per hour. 
Here d has been taken as 4.9 ft. Using this latter formula, the 
following table for the superelevation of the outer rail has been con- 
structed : 

Elevation Qi) in Feet for Given Radius in Feet 



VELOCITY 
MILES PER 


5730 


2865 


1910 


1632 


1146 


955 


HOUR 














20 


.02 


.05 


.07 


.09 


.12 


.14 


30 


.05 


.10 


.16 


.21 


.26 


.31 


40 


.09 


.19 


. .28 


.37 


.46 


.56 


50 


.15 


.29 


.44 


.58 


.73 


.87 


60 


.21 


.42 


.63 


.84 


1.04 


1.25 



88. Simple Circular Pendulum. — The simple circular 
pendulum consists of a weight Gr suspended by a string 
without weight, of length ?, in such a way that it is free 
to move in a circle in a vertical plane due to the action of 
gravity (see Fig. 111). Let B be such a position of the 



CURVILINEAR MOTION 



149 



pendulum that its height above the liorizontal is h, and 
c any other position designated by the coordinates x 
and y. Let the weight 
be (7 and the tension 
in the string T, These 
are the only forces act- 
ing on the body O, 
The only forces that 
can produce motion in 
the circle are those 
that are tangent to the 
circle OB. The force 
T is normal to the 
circle and so has no 
tangential component. 
The force Cr has a tangential component — Gr sin a. The 
equation of motion is, therefore, 

vdv = ads = — ^ sin ads^ 

V 

„,-u^ arcT^?^) s . 

wneie a = = Avhere s denotes distance along the 

curve. 

This equation, as it stands, leads to a complicated rela- 
tion between distance and time. If, however, the angle 
a is sufficiently small, so that we may replace sin a by a, we 
may write 

I vdv = ^^ ( sds. 
Integrating with respect to s, we liave 




.'=f.,^-j«». 



150 APPLIED MECHANICS FOB ENGINEERS 

where v = 0, s= S;,, where s^ is length of curve OB. 
Solving for the velocity, we have 



or ar^r^' ^' 

which may be put in the form 

where c is a constant of integration. 
This equation may be written, 



s = Sj, cos 



[VfO-o]; 



it represents the relation between distance and time. 
When s = Sf^^ t = t^^ so that 

[Vf0.-O]=0; 

therefore, c = t^^. 

It is evident that s is a periodic function of the time, 
and that it repeats itself at intervals of time t^ such that 

n 

^9 
This value t^^ represents the time taken by the body from 
leaving the position B until its return. One half of this 
value P- 

h g 

is designated as the period of vibration. 



CURVILINEAR MOTION 151 

In general, when the angle is not small, the equation 
vdv = — g sin ads becomes, since ds sin a = dy^ 

Integrating, we have v'^=: —2gy =2g(^h — y). 

It will be seen that this result is the same as if the body 
had fallen freely through a height h — y (see Art. 74). 
The value for v is evidently true whatever be the vertical 
curve in which the body moves, providing the only forces 
acting on the body are the force of gravity and another 
force normal to the curve. The foregoing fact leads to 
the statement, in descending along any curve without fric- 
tion^ from a height h to any other height y a body will have 
the same velocity as if it fell freely through the height h — y. 
This fact is often made use of in mechanical problems. 

The summation of forces normal to the path gives 



T— Gr cos a = 



I ' 

Mv^ 



and, therefore, T= Gr cos a + 



r' 



which gives a value for the tension in the cord. Since v 
is greatest when a = 0, it is seen that the greatest tension 
in the cord occurs when the pendulum is vertical. 

If now we make use of the fact that the body moves in 
a circle whose equation is x^+ y^— 2 ry = 0, and remem- 
ber that ds^ = dx^ + dy^ for any curve, we may write 



v^ = 



dt 



152 APPLIED MECHANICS FOR ENGINEERS 

and, therefore, f — ) = ^^^ =^1g(h- y), 



or dt 



-\/dx^ + d^/^ 



But from the equation of the curve 

so that ^^2^ 0/-O^y 
and rf^ = — — ^ 



The integral of the expression on the right-hand side of 
the equation is not expressible in terms of ordinary alge- 
braic or trigonometric functions, but must be expressed in 
terms of the elliptic functions. The student may not be 
familiar with such functions, so that we shall express it 
approximately by means of an infinite series. This series 
will be sufficiently rapidly convergent if the radius of the 
circle is large and the distance OB is small. Using the 
minus sign in the numerator, since ^ is a decreasing func- 
tion of s, we may write 



'^ jr rh dy ^ _ xV^ 
^yJo L 2V2r; 2 4t\2rJ 



\2 
+ 



dy 



^9 



2r \2--i:J\2rJ VI-l-aJK^rJ J 



^-&i-AMm'-(^Mh^-^- 



CURVILINEAR MOTION 



153 



For very small values of h we may neglect the terms con- 
taining h. The result then becomes 

since r = Z, and this is the same result that was obtained 
before. 

This means that for small values of a, not greater than 
4°, the time of vibration of a simple circular pendulum is 
a constant ; that is, the oscillations are isochronal. 

It is seen that the time of vibration of a simple pendu- 
lum varies as the square root of its length, for any locality 
on the earth. In order to get a pendulum that will beat 
seconds it is necessary to place ^=1. Knowing the value 
of g for the locality, the proper length may be determined. 
If w^e measure the length of a pendulum and its period 
we may calculate the value for^ for any locality. This is 
the easiest and most accurate way of determining g. 




Fig. 112 

Problem 128. The centrifugal railway (Fig'. 112), or ''loop the 
loop," is a conuiion example of a simple circular pendulum, where 
the effect of the string is replaced by a track. If we neglect friction, 
the only forces acting on the car are the force of gravity and the 



154 



APPLIED MECHANICS FOB ENGINEERS 



normal pressure of the track. Suppose the car starts from rest at a 
height, h. What must be the relation between h and h', so that the car 
will pass the point A without leaving the track. 

Hint. The velocity at the lowest point, v^= 2 gh, is the same as the 
velocity with which the car comes down. The centrifugal force must 
be great enough at A to overcome G, the w^eight of the car (h = |A'). 

Problem 129. In the simple pendulum find the value of y in 
terms of h for which the tension in the string is the same as when the 
pendulum hangs at rest. 

Problem 130. A pendulum vibrates seconds at a certain place and 
at another place it makes 60 more vibrations in 12 hours. Compare 
the values of g for the two places. 

89. Cycloidal Pendulum. — It has been found that a pen- 
dulum may be obtained whose period of vibration is 




Fig. 113 



constant by allowing the string to wrap itself around a 
cycloid as shown in Fig. 113. The pendulum hangs from 
the point A, AB and AO are cycloidal guides around 



CURVILINEAR MOTION 155 

which the string wraps as the pendulum swings. This 
causes the length of the pendulum to continually change 
and the pendulum ''bob" to move in another cycloidal 
curve COB, The equation of this curve referred to the 
axes X and ?/ is 

a; = -vers [-l^ + ^j^-y\ 

In Art. 88 it was seen that v^=^2g(h — y^ represented 
the velocity of a body moving in a vertical curve when 
only the force of gravity and a force normal to the path 
of the curve acted. We may make use of the equation in 
this case, since the same conditions exist. We may write 

^^ ^dx^ + dip' . c?s 

at = — — ==^, since ^; = --• 
^2g{h-y^ dt 

From the equation of the curve, we find 



r 



so that Cdt = a/-^ r (- jT—^ j 

taking the negative sign, since ^ is a decreasing function 
of s. 

Therefore t = \~- vers~^-/ =7r\/ — -. 
^-igl A Jo yi4g 

The whole time of vibration is twice this value, so that 
the time of vibration 



156 APPLIED MECHANICS FOR ENGINEERS 

This expression is independent of A, so that all vibrations 
are made in the same time. The motion is therefore 
isochronal. 

Problem 131. A body of mass M slides from rest down a cycloid 
from the position 5 (Fig. 113) without friction. What is its velocity 

when ^ = -? Show that this is its maximum velocity. 

Problem 132. Find the position of a cycloidal pendulum w^here 
the tension in the string is greatest. What is the velocity of the bob 
at the point O (Fig. 113)? 

90. Motion of Projectile in Vacuo. — A method, slightly 
different from the preceding, of dealing with a problem 

of curvilinear 
motion, is il- 
lustrated in 
the present ar- 
ticle. It is de- 
X sired to find 
the path taken 
by a body pro- 
jected with a velocity v^ at an angle of elevation a, when 
the resistance of the air is neglected (see Fig. 114). In 
this case, since there is no horizontal force acting on the 
body a^, = 0, so that, 

and -^r- = constant = v\ cos a, 

at ^ 

therefore, x = Vq cos a {t). 

In a similar way we know that the vertical acceleration 
ay= — ^, since the only force acting is Gr. 




Fig. lU 



CURVILINEAR MOTION 157 

Then, §=-, 

and -f^ = — gt + constant. 

at 

This equation may be rewritten 

Vy= — gt-\- constant. 

To determine this constant of integration, we put ^ = 0, 

and Vy = v^y = v^ sin a; 

therefore ~J~^ ~9^ + '^o sin a 

and y = — }^ gt'^ 4- v^ sin a (f). 

Eliminating t between the equations in x and y, we get 

2/ = i:c tan a - ^^-^f — r- 
2 vi cos^ a 



as the equation of the path of the projectile. This is evi- 
dently a parabola, with its axis vertical and its vertex at A, 
Range, To find the range or horizontal distance d we 
put «/ = : then x = Q and x = d^ 

,1 , , t^osin2a 

so that d = — 

9 

From this it is clear that the greatest range is given when 

a == 45°, since then d = -^^ 

9 

The Grreatest Height. The greatest height to which the 
projectile will rise is found by putting x = -^^ — ^^ in 
the equation of the curve and solving for g. Tliis gives 



158 APPLIED MECHANICS FOR ENGINEERS 

and the angle that gives the greatest height is a = 90°. 

For this case h = ^ . This is the case that has already 

been considered under the head of a body projected verti- 
cally upward. 

91. Body projected up an Inclined Plane. — If the body is 
projected up an inclined plane, making an angle yS(/3<a) 
with the horizontal and passing through the point (see 
Fig. 114), we desire to find the point at which the pro- 
jectile will strike the plane. For any point in the plane 
we have y ^x tan y8. The point where this plane cuts the 
parabolic path of the projectile is given by the equations 

2 ^0 cos a sin (fa — /3) 

2 v^ cos a tan ^ sin (a — /?) 
^1"" ^cosyS 

The range on the plane 

2vl cos a sin (a — /3) 
g cos^yS 

Problem 133. The initial velocity vq is the same as that of a body 
falling freely from the directrix of the parabolic path to the point 
on the curve. Show that the velocity of the body at any point on the 
curve is the same as would be acquired in falling freely from the 
directrix to that point. 

Problem 134. A fire hose delivers water with a nozzle velocity vo, 
at an angle of elevation a. How high up on a vertical wall, situated 
at a distance c/' from the nozzle, will tlie water be thrown? It should 



CURVILINEAR MOTION 



159 



be said that water thrown from a nozzle in a non-resisting medium 
takes a parabolic path and follows the same laws as projectiles. 

Problem 135. What must be the nozzle velocity of water thrown 
upon a burning building, 200 ft. high, the angle of elevation of the 
curve being 60° ? 

Problem 136. The muzzle velocity of a gun is 500 ft. per second. 
Find its greatest range when stationed on the side of a hill which 
makes an angle of 10° with the horizontal : (a) up the hill, (h) down 
the hill. If the hillside is a plane, the area commanded by the gun 
is an ellipse, of which the gun is a focus. 

Problem 137. A ball whose weight is 64.4 lb., shown in Fig. 
115, starts from rest at A and rolls without friction in a circular path 




/ ^ G =64.4 



LBS. 



/ 






Fig. 115 



to the point B, where it is projected from the circular path horizontally. 
Find (a) the velocity at B, (b) the equation of its path after leaving 
B, and (c) the distance d from a vertical through B, where it strikes a 
horizontal 10 ft. below B. 



160 



APPLIED MECHANICS FOB ENGINEERS 



Problem 138. If the body in Problem 137 had moved along a 
straight line from A to B and was then projected, find, as in the 
preceding problem, (a), (h), and (c). 

Problem 139. A body whose weight is 12 lb. swings as a circular 
pendulum, as shown in Fig. 116, from A to B, w^hen the string breaks. 

Find (a) the velocity at 
B, (b) the equation of 
its path after leaving 
B, and (c) the distance 
d where it strikes a hori- 
zontal 5 ft. below B. 




/ 



d 



/ to 



'12 LBS. 



Ci 



Fig. IIG 



Problem 140. A 

ball whose weight is 
32.2 lb. starts from rest 
at A on the top of a 
sphere (Fig. 117), and 
rolls w^ithout friction to 
the point B, where it 
leaves the surface. Lo- 
cate the point B. Find 
also (a) the angle of 
projection a, (h) the 
equation of the path of 



the body after leaving the sphere, and (c) the distance d where it 
strikes the horizontal. 

Problem 141. The muzzle velocity of a gun situated at a height 
of 300 ft. above a horizontal plane is 2000 ft. per second. Find the 
area of plane covered by the gun. 

Problem 142. The fly wheel shown in Fig. 81 ^'runs wild " and 
the rim breaks into six equal parts, free from the arms, when going at 
the speed of 300 revolutions per second. The path of the pieces being 
unimpeded, find the greatest height that could be reached by either 
piece and the greatest horizontal distance attainable. 

^^ 92. Motion of Projectile in Resisting Medium. It was 
found by Rollins and others (see Encyclopa3dia Britannica 



CURVILINEAR MOTION 



161 



— "Gunnery'') that 
the formula for pro- 
jectiles in vacuo did 
not hold when the 
projectile moved in 
the atmosphere. That 
is, that the path fol- 
lowed by the projec- 
tile was not parabolic, 
but on account of the 
resistance of the at- 
mosphere the range 
was much less 
than that given 
by the parabola. 

A formula constructed by Helie, empirically modifying 
the parabolic formula, is 




Fig. 117 



y = X tan a — 



gx^ 



2 cos^ a 



kx 



d^ 



where k = 0.0000000458 — , d being the diameter of the 

w 

projectile in inches, and w its weight in pounds. This gives 

the simplest formula for roughly constructing a range table. 

Professor Bashforth of Woolrich found, from a series of 
experiments made by him, that for velocities between 900 
and 1100 ft. per second the resistance varied as v^^ for 
velocities between 1100 and 1350 ft. per second the re- 
sistance varied as v^^ and for velocities above 1350 ft. per 
second the resistance varied as v^. 

In addition to the resistance of the air other factors tend 
to change the path of the projectile from the parabolic 



102 APPLIED MECHANICS FOR ENGINEERS 

form, viz. the velocity of the wind and the rotation of the 
projectile itself. Most projectiles are given a right-handed 
rotation, and this causes them to bear away to the right 
upon leaving the gun. This is called drift. Correction 
is made for drift and wind velocity upon firing. 

If the resistance of the air varies as the velocity, say it 

equals kv^ then kv^ = k —- and kvy = k -^^ 
^ "" dt "" dt 

so that 

^ ^ df^ dt' dt^ dt ^* 

Integrating, and remembering that when ^ = 0, 

v^ = v^ cos a, Vy =v^ sin a, 
we have 

/^ox dx _j,t dy In 

(2) ..= - = .„cos«..- v, = f^ = -i-g 

+ {kvQ sin a + ^)^"^'^], 
and therefore, since, when ^ == 0, a; = 0, and 2/ = 0, 

(3) a; = ^ cos a (1 - ^-^'), 

k 

(4) y=.^t-{- ' ^^ ^(1 - e ^y 

Eliminating ^, the equation of the curve is 

/rx kvi) sina + a , q ^ ^n cos a—kx 

(5) 2/ = — 7 ^^ ^ + 7^1og-^ . 

kv^ cos a k^ Vq cos a 

93. Path of Projectile Small Angle of Elevation. — When 
the resistance varies as the square of the velocity, the 
complete determination of the path of the projectile is 
mathematically difficult. In what follows, tlie angle of 



CURVILINEAR MOTION 163 

elevation has been assumed small so that powers of ^ 

dx 

higher than the first have been neglected. Then ds = dx 
and s = x. Let the resistance equal kv'^= kl-—j . Then 

.^. d^x __ ndsdx^ 

^ ^ dfi" Jtdt' 

d?y 7 ds dy 

Equation (1) may be put in the form 

d{^\ 
\dtJ ___^,ds 

dx dt 

which gives, upon integrating, 

dx 

(3) , dt 

log _ = — ^s. 



Vq cos a 



dx 

'dt 



Cm I 

Since the initial value of -77 is v^ cos a, that is, when ^ = 0, 



dx 

— • = -y^ cos a, 

dt ^ 



Equation (3) may be written, 
(4) — = 2;^= t; cos a-^ 



Multiplying (1) by dy and (2) by dx and subtracting 
(2) from (1), we get 



164 APPLIED MECHANICS FOR ENGINEERS 



(5) ^y^^^Il^^ = -gdx. 



From (4) and (5) we have 



d^ydx-dHdy _ ^ ^ _ _ 9 ^,^^^^ 



ddi? dx 7.2 QQg2 ^ 



and since 8 = 2:, 

(6) d^= '-L e^'^^dx. 

Integrating, and remembering that when 2: = 0, -^ = tana, 

dx 

we have 

(7) ^ - tan a = 3_ (^2^- _ 1) . 

therefore, 

(8) ^=:,tan«+--f^-^ ,„/ , (.2^--l). 

2 A:?;^ cos"^ a 4 A:^y^ cos^ a 

If ^2^"^ be expanded in a series, this may be expressed 
approximately as follows, 

(9) y = xi'i\A\a ^^ f ^. • • • 

It is seen that if the third and following terms be neglected, 
the equation is that of the projectile in vacuo (Art. 90). 

Problem 143. Find the range, greatest height, and time of flight, 
from llelie's equation (Art. 92) ; Equation 5, Art. 92 ; and Equation 9 
of the present article. 

Problem 144. Compare the values obtained in the preceding 
problem with similar values obtained for the case of motion in vacuo, 
taking a = 45° and vq = 1000 ft. per second. In each case take 

k= .0000000458 — , where J = 6 in. and w; = 150 lb. 

w 



CURVILINEAR MOTION 165 

Problem 145. Find the angle of elevation a for each of the cases 
in preceding problem in order to strike a point 200 ft. high, distant 
1000 ft. Take r, = 1000 ft. per second. 

Problem 146. A locomotive weighing 175 tons moves in an 800 
ft. curve with a velocity of 40 mi. per hour. Find the horizontal pres- 
sure on the rails, if they are on the same horizontal. 

Problem 147. If the velocity of the earth was 18 times what it 
actually is, show that the force of gravity would not be sufficient to 
keep bodies on the earth near the equator. Take the radius of the 
earth as 4000 mi., and assuming the above conditions, find at what 
latitude the body would just remain on the earth. 

Problem 148. The weight of a chandelier is 300 lb., and the dis- 
tance of its center of gravity from the ceiling is 16 ft. Neglecting the 
weight of the supporting chain, find how much the tension in the 
chain will be increased if the chandelier is set swinging through an 
angle of 2°, measured at the ceiling. 

Problem 149. A pail containing 5 lb. of water is caused to sw^ing 
in a vertical circle at the end of a string 3 ft. long. Find the velocity 
of the pail at the highest point so that the w^ater will remain in the 
pail. Find also the velocity of the pail at the lowest point. 

94. Motion in Twisted Curve. — When the motion of a 
body is in a twisted curve, it is convenient to take account 
of its motion relative to three rectangular axes, x, ?/, and z. 
Let a, yS, 7 be the direction angles of the tangent line to 
the curve, and X, //., v the direction angles of the resultant 
force. We may then write for the velocity, 



v^ = v cos a = 



dx . 
lit' 



Vy = v cos /3 = —^ I 



dz . 
v^ = v cos 7 = — » 

dt 



166 APPLIED MECHANICS FOR ENGINEERS 



.= vgTiiT7i=v('|)V(;IT+(IT- 



MJ \dtj \dt 

and for the acceleration, since 

dx __ dx ds Q^^Q 
dt ds dt 

cPx c^s dx fds\^ d^x 
a^ = a cos A, = -r— r = ———- + 



dt^ dt^ds \dtj ds^ 
a, = acos/.=--^ = — -^+ - ^• 



dfi df-ds \dtj dt^ 

dh dH dz fds\^ d'^z 
«. = acos. = ^=^,^^ + (^j 5^- 






dt^J \dty \dt^J' 

since — 5 — ^? — - are the direction cosines of the tangent 
ds ds ds 

line and p — --> p— ^, andp — - are the direction cosines of 
ds^ ds^ ds^ 

the principal normal. 

From the above equations it will be seen that the result- 
ant acceleration a may be resolved into tangential and 
normal components 

at = —- and a^ = —-> 
dt p 

just as was done in the case of motion in a plane curve. 
In this case the normal is the principal normal and the 
radius p is the radius of absolute curvature. 

As an illustration of motion in a twisted curve consider 
the motion in a helix. The helix may be considered as 



CURVILINEAR MOTION 



167 



generated by the end of a line that moves with uniform 
velocity along the line OZ (Fig. 118). The edge of the 
thread of a screw is such a twisted curve. Let the curve 




Fig. 118 

be given by Fig. 118, and letP be any point having coor- 
dinates x^ y, and z. 

Then x = r cos c^, 

^=rsin^, 



168 APPLIED MECHANICS FOB ENGINEERS 

will represent the curve. It follows that 

t; = — r sin d) -^ = — ro) sin </> ; 

V =r cos 6-^ = rco cos (f>; 
^ at 

_ k cl(f> _cok 

-2 is the angular velocity of the point with respect to z; 

^^ I p" 

represent it by co (see Art. 95), so that v = o)^ r^ -] 

= constant (^k is an arbitrary constant that determines the 
pitch of the helix). The velocity of a point moving in 
such a curve is constant since co is constant. The accel- 
eration a^ is therefore zero. 
We may also write 

a^== r cosrf)-^ = —rco cos 6; 
^ ^ dt 

ay — r sin </> -^ = — rco sin ^. 

a. = 0. 



Therefore a = V a| + a^ + a^ = ray. 
That is, the acceleration in the direction of the resultant 
force is equal to (or and the accelerating force is equal to 

The fact of zero tangential accelerations has made this 
curve very useful. In many cases the helical surface 
formed by the revolving line has been made use of to 
send packages from upper floors of commercial establish- 
ments to the lower floors. Since a^ = 0, the packages 
move down with uniform motion. The helicoid is in- 
closed in a tube with convenient openings for the insertion 
of packages. 



CHAPTER XI 

ROTARY MOTION 

95. Angular Velocity. — In Art. 71 linear velocity was 
defined as the rate of motion, and it was stated that it 
might be expressed as the ratio of distance to time or 
the rate of change of linear distance to time. The 
simplest case of rotating bodies is seen in uniform rotation 
about a fixed axis. The angular velocity is defined as 
the ratio of angular distance to time. Let the angular 
distance (measured in radians) be represented by a and 
the angular velocity by co. Then for uniform velocity 

» a 

and for variable velocity 

dt 

Angular velocity involves a magnitude and a direction, 
and may, therefore, be represented by an arrow (see 
Fig. 119), the length of the arrow representing the magni- 
tude and drawn perpendicular to the plane of motion 
such that if you look along the arrow, from its point, 
the motion appears positive or negative ; positive if coun- 
ter-clockwise and negative if clockwise. 

Velocity arrows may be compounded into a resultant 
or resolved into components in the same wa}' tliat force 
arrows were treated. For example, in Fig. 119, the 

169 



170 



APPLIED MECHANICS FOR ENGINEERS 



angular velocity of a body at any instant is represented 
by CO. Then the angular velocity with respect to two rec- 




•33— 



Fig. 119 



tangular axes x and y will be represented hj co^ = co cos \ 
and (Oy = co sin X, so that co^ = toj + co^. 

In a similar way if a body has an angular velocity 
CO about an axis making angles X, /x, and v with the x^ y^ 
and z axes, respectively, the component velocities along 
the axes will be given by 



ft)^ = ft) cos X, o)^ == ft) cos /i, ft)^ = ft) cos V 



so that 



a)2=z:«2 +0,2 +0)2. 



96. Angular Acceleration. — Angular acceleration, which 
we shall represent by ^, may be defined as the rate of 
change of angular velocity, so that if 






ROTAIIY MOTION 171 

From the preceding article and the definition of angular 
acceleration we may write 

'~~dt' ^~ dt ' '"'dt' 

The linear velocity and linear acceleration of a point 
of a rotating body may be determined in terms of the 
angular velocity and angular acceleration. Assume that 
for the instant under consideration the point is moving 
in the arc of a circle of radius p, over an arc ap, then 

pda 1 pd'^a 

i; = ^ = cop and at = ^^ = Op. 

dt dfi 

It will be seen that v in this case is the velocity along a 
tangent to the path at the point P. 

/- 97. Angular Acceleration Constant. — In Art. 73 we found 
that when the linear acceleration was constant, the equa- 
tions of motion reduced to a simple form. In a similar 
way if is constant, and the axis of rotation fixed, we have 

« = «o + 0^ ; 

(lidia = Met ; 

<^= — ^— ^; 

20 

where (o^ is the constant initial angular velocity. 

The expression for linear velocity and linear accelera- 
tion of any point P of the body becomes in this case 
V = a^r and at = ^r, where r is the distance of P from the 
axis of rotation. 



172 APPLIED MECHANICS FOR ENGINEERS 

Problem 150. A fly wheel making 100 revolutions per minute is 
brought to rest in 2 min. Find the angular acceleration and 
the angular distance a passed over before coming to rest. 

Problem 151. A fly wheel is at rest, and it is desired to bring it 
to a velocity of 300 radians per minute in J min. Find the accelera- 
tion 6 necessary and the number of revolutions required. What is 
the velocity w at the end of 10 sec. ? 

Problem 152. Suppose the fly wheel in Problem 151 to be 6 ft. 
in diameter. After arriving at the desired angular velocity, what is 
the tangential velocity of a point on the rim? What has been the 
tangential acceleration of this point, if constant ? 

98. Variable Acceleration. — In case is not constant, 
its law of variation must be given so that the equations 
of motion may be worked out. As an illustration suppose 
that a body moves in such a way that the angular accelera- 
tion 6 varies as the angular distance a. Let ^ = — - ^cc, 
then from the equation codco = dda^ we get (odco = — kada. 
Taking the limits of (o as (o^ and c», and the limits of a, 
and a, and of t^ and t^ we have 



Jcodco = — k ) ada ; 



therefore co = Va)2 — ka^. 

Integrating again, 

da 



which gives 



J"^"^ da r 

Va)2 — ka^ *^Q 





1 . _iV^a 
— - sin ^ = t^ 

■\/k «o 



or -^ sin -^u = <^- 

-\k 



ROTARY MOTION 



173 



This last equation sliows tliat a is a periodic function 
of the time; the motion is vibratory. Referring to Art. 
79, it is seen that the motion is harmonic. In fact, if we 
substitute co^ = ij^p and a =^/o, we have exactly the same 
equation as was obtained in Art. 79. This example ap- 
plies to the motion of a simple pendulum, considering 
it as rotating about the point of support. 

Problem 153. The balance wheel of a watch goes backward and 
forward in | sec. The angle through which it turns is 1S(F; find 
the greatest angular acceleration and the greatest angular velocity. 

Problem 154. Assume the angular acceleration varies inversely 
as the square of the angular distance; find the relation between oj and 
a, and t and a. 

99. Combined Rotation and Translation. — The angular 
acceleration of a body may 
be resolved into its tan- 
gential and normal com- 
ponents at = 6p and «^ = 

— = ofip. If now the body 

has in addition to its rota- 
tion, a translation, the total 
acceleration of any point 
P will be given by the 
components 6p^ od^p^ and %. 
In Fig. 120 the body is sup- 
posed to have an axis of ro- 
tation perpendicular to the 
paper, and a translation parallel to ox. Let the angle that 

PO makes with x be yS, so that cos (3 = - and sin ^ = --• 

P P ' 



ar^ 




Fig. 120 



174 APPLIED MECHANICS FOR ENGINEERS 

Writing the x and 7/ components of the acceleration, we 
have 



a^ = — co^x 



Oy -aj 



ay= — ofiy + Ox. 
The tangejitial and normal components of a are, from 

^'^' ^^^' fv\ fx\ 

at = 6p + aJ^\ a^ = co^p + aJ-V 

As an illustration of combined rotation and translation 
consider the case of a wheel of radius r rolling in a 
straight horizontal track. Let the acceleration of transla- 
tion of the center be a^, and the angular velocity of the 
wheel, about the center, be w, and the corresponding 
angular acceleration 6. 
Tl^en 

are the tangential and normal accelerations of a point on 
the rim situated at the top of the wheel. 

Problem 155. A locomotive drive wheel 6 ft. in diameter 
rolls along a level track. Find the greatest tangential acceleration 
and the greatest normal acceleration of any point on the tread, 
(a) when the velocity v with which the w^heel moves along the track is 
60 mi. per hour, (b) when the engine is slowing down uniformly 
and has a velocity of 30 mi. per hour at the end of 3 min., 
(c) when the engine is starting up uniformly and has a velocity of 
30 mi. per hour at the end of 5 min. 

Problem 156. A cylinder, diameter d, rolls from rest down an 
inclined plane, inclined at an angle <^ wdth the horizontal. What is 
the greatest normal and greatest tangential acceleration of any point 
on its circumference? I^eglect friction. 



BOTAllY MOTION 175 

100. Rotation in General. — It lias been shown in Art. 
36 that any system of forces acting upon a rigid body 
may be reduced to a single force and a single couple 
whose plane is perpendicular to the line of action of the 
single force. That is, the most complicated cases of rota- 
tion consist of an instantaneous translation combined with 
an instantaneous rotation at right angles to the translation. 

Bodies projected into the air while rotating have been 
mentioned in Art. 92. The projectile rotating about an 
axis is projected in the direction of the axis. If no forces 
acted upon it after leaving the gun, it could move in a 
straight line. It is, however, acted upon by gravity, which 
causes it to take a somewhat parabolic path. The resist- 
ance of the air causes the projectile to drift. 

This action of the projectile will probably be most 
easily explained by a consideration of the motion of a base- 
ball. The modern pitcher when he throws the ball gives 
it also a motion of rotation. The force of gravity causes 
the ball to take a path somewhat parabolic and the resist- 
ance of the air, due to the rotation, causes the ball to de- 




:^ — > 



Fig. 121 



176 APPLIED MECHANICS FOR ENGINEERS 

fleet from the plane in which it started. The combination 
of the two deflecting forces makes the path of the ball a 
twisted curve. Different speeds and directions of rota- 
tion and different speeds of translation give great variety 
to the curves produced. The action of the baseball will 
be best understood by referring to Fig. 121. Let the 
baseball have an initial angular velocity co and an initial 
linear velocity v in the directions shown. The rotation 
of the ball causes the air to be more dense at J. than at jB, 
so that the ball is pushed constantly from A to B, This 
action causes it to deviate from the plane in which it 
initially moved and to take the path indicated by c. 
As stated above, this action in the case of a projectile 
is known as drift. 



CHAPTER XII 

DYNAMICS OF MACHINERY 

101. Statement of D'Alembert's Principle. — A body may 
be considered as made up of a collection of individual 
particles held together by forces acting between them. 
The motion of a body concerns the motion of its individ- 
ual particles. We have seen that in dealing with such 
problems as the motion of a pendulum it was necessary to 
consider the body as concentrated at its center of gravity; 
that is, to consider it as a material point. The principle 
due to D'Alembert makes the consideration of the motion 
of bodies an easy matter. Consider a body in motion due 
to the application of certain external forces or impressed 
forces. Instead of thinking of the motion as being pro- 
duced by such impressed forces, imagine the body divided 
into its individual particles and imagine each of the parti- 
cles acted upon by such a force as would give it the same 
motion it has due to the impressed forces. These forces 
acting upon the individual particles are called tlie effective 
forces, D'Alembert's Principle, then, states that the im- 
pressed forces ivill he in equilibrium with the reversed effec- 
tive forces. 

It must be seen by the student that the principle does 

not deal with the forces acting between the particles of a 

body; these are considered as being in equilibrium among 

themselves. We shall see in what follows that this 

N 177 



178 



APPLIED MECHANICS FOR ENGINEERS 



principle, by assuming a system of effective forces acting 
upon the particle, enables us in many cases to apply the 
principles of equilibrium as developed and used in the 
subject of statics. 



102. Simple Translation of a Rigid Body. — The principle 
of D'Alembert will be best understood by applying it to 

•' the consideration of 
the simpler motions 
of a rigid body. Let 
us consider the body, 
Fig. 121 a, and let us 
assume that it has 
simple translation 
parallel to x due to 
the action of certain 
impressed forces P., 




Fig. 121a 



^ P^, Pg, etc., making 

angles oci, aoj ^3' ^^<^-' 
with X, It is seen at once that only the components 
of Pj, P2, P3, etc., parallel to x have any part in produc- 
ing motion in that direction. We may say, then, that 
the impressed forces are Pj cos a^, P^ cos a^-, P3 cos ccg, etc., 
and that these produce an acceleration a in the direction 
indicated. 

Imagine the body now divided into small particles each 
of mass dM^ and assume that the system of forces produc- 
ing the motion of the body consists of a small force dM . a 
acting on each particle. D'Alembert's Principle then 
states that these forces reversed are in equilibrium with 
the impressed forces. We have, then, ^x =0, 



DYNAMICS OF MACHINERY 179 

or PjCOS ai + 1^2^^^ «2 + ^3C0S a^ + etc. — ^tlM • a = 0, 
or P-^cos ai + P^^os a^ + -Ps^^os a^ + etc. = ^dM • a. 

But since motion is parallel to x^ it is evident that 
Pjcos ai + P'f'^^ a^ + Pz^o^ a^ + etc. = Resultant Force = 72. 

Therefore, for continuous bodies, 

since a is the same for every particle of the body. Con- 
sider each particle at a distance y from x and let d be the 
distance of R from x ; then taking moments with respect 
to an axis through x and perpendicular to it, we have 

Rd — a \ydM = ayM^ 

where y is the distance of the center of gravity of the body 
from X (Art. 22). Dividing through by R^ we find, 

that is, the resultant force passes through the center of 
gravity of the body. 

103. Simple Rotation of a Rigid Body. — We shall now 
apply D'Alembert's Principle to the case of a rigid body 
rotatinof about a fixed axis. Let B in Fisr. 122 be the 
body, and imagine it rotating in the direction indicated 
about an axis through perpendicular to the paper. Sup- 
pose the rotation due to the action of forces P^, P^^ Pg, P^, 
etc., making angles a^, ^^, 7^ a^. ^^^ j^' ^^^"> ^^'i^l^ ^ ^^^ of 
axes x^ 2/, ^, with origin at 0. It is evident that only the 
components of the forces P^, P^^ P3, P^, etc., parallel to 



180 



APPLIED MECHANICS FOB ENGINEERS 



the xz-'plaue^ Avill have any part in producing rotation. 

Call these projections P^, P^^ ^s^ P['> ^^^-5 ^l^^J ^^'^ ^^^ 
impressed forces 
for the motion con- 
sidered. The dis- 
tances of the lines 
of action of these 
forces from may 
be represented by 

(Jv-ttj Ctey^ ^Q^ ^49 etc. 

Now consider 
the effective 
forces. Imagine 
the body made up 
of individual par- 
ticles c^iltf situated 
at a distance p 
from 0. Eachc^iHf 
is acted upon by 
a force d3Iat=d3I6p, These are the effective forces. 
Equating the moments of these forces, reversed, to the 
moments of the impressed forces, we have 

2 {Pi'di + r2.'d2 + jPs'^s + etc.) = (cUI-ep'p = 6 Cp'^d3I= 07, 

since j p'^dM gives the moment of inertia of B with respect 
to 0. (See Art. 37.) 

That is, when a body rotates about a fixed axis, the sum of 
the moments of the impressed forces in the jjlane of rotation 
equals 61. 

It is evident that any one of the forces P[^ P^, P^, F^, 




Fig. 122 



DYNAMICS OF MACHINERY 



181 



etc., may be such as to offer a resistance to the indicated 
motion of the body. In such a case the sign of its mo- 
ment would be changed. 



104. Reactions of Supports ; Rotating Body. — It has 
just been shown that one equation is sufficient to give 
the motion of a 
rigid body about 
u fixed axis. It 
is necessary, how- 
ever, in order to 
determine the re- 
actions of the 
supports, to use 
other equations. 
Consider the 
body B with its 
axis vertical, as 
shown in Fig. 
123, and let the 
rotation take 
place as indi- 
cated due to the 
action of the 
forces P^, P. 



-£3, x'^^, etc. 



2' 

Let 




Fig. 123 



P^, Py^ and P^ be the reaction of the supports on the axis 
at 0, and P^ and P^ the reactions of the support at A on 
the axis. 

The effective forces acting upon a particle of the body 
B are shown in Fig. 124. Let dMhe its mass and replace 



182 



APPLIED MECHANICS FOR ENGINEERS 



the resultant force acting on d3f by its tangential and 
normal components and call them cZ^ and JiV respectively. 

It has been shown (Art. 

86) that the normal force 




equals 



3Ii 



and the tan- 



gential force equals Ma^^ 
and that v = cop and a^ = 
Op (Art. 96). We have 
then 

dN = fOJa)2p and d T= eZ JX0p, 

X and it is seen that both 
dJV and dT may be re- 
solved along each of the 
axes x^ y, and z for every 
dM of the body. 

From what has been 
said it is evident that the sum of the impressed forces 
along the a;-axis equals the sum of the reversed effective 
forces along the same axis. Calling the impressed forces 
i and the effective forces ^, we may write 



2.x i — ^x^^ 



X mom^^J. = 2 mom^^, 
S mom.,, = S mom 



w 



ey^ 



S niom^v^ = S mom^ 



That is, the components of the impressed forces along 
each of the three axes x^ y, and z equal the components 
of the reversed effective forces along these axes and the 
moment of the impressed forces with respect to the three 



DYNAMICS OF MACHINERY 183 

axes x^ ?/, and z equals the moment of the reversed effective 
forces with respect to these axes. 

Writing down these six conditions, we have 

^Xi = — (dlVcos (f> + (dTiiin <^, 

2y^- = — (dN'sm (f) — \dTcos(f>^ 

2 mom^-^ ~ "" J ^^^^'^'^ • ^ + J dT cos cf) • 2, 

S mom,-^ = \dJVcos (f) - z+ idTsin (f) • ^, 

Smom^2= J dT' p. 

Substituting the values of tZiVand dTin the expressions 
on the right-hand side, we have 

^dJVcos (f) = w2 Jp COS (j)d3f= afi^xdM= - ay^Mx, 
^dNsin (t> =co^^ydM= - co'^M^, 
Jc?iV^sin ^ . 2 = ofl^ijzdM, 
JrfiV^cos (/) • ^ = &)2 Ja:^t7il!f, 

f cZrsin ^ = ^epdMsm cf> = e^ydM=e3Iy, 
(dTco^ (\> = e^xdM= OMx, 
jdTsm(f>'Z = 6^f/zdM, 
^dTcos<f>'Z = e^xzdM. 



184 APPLIED MECHANICS FOR ENGINEERS 

The six general equations therefore reduce to the form 
^Xi = QMy - 0)2 JX^, (^1) 

^y. = -r 0315c - a)2ilf ^ (2) 

S^i=0; (3) 

2 mom^^ = - «2 (yzdM + (xzdM, (4) 

S mom ,.^ = 0)2 fxi^i^JJr -f ^(yzdM, (5) 

2 mom^^ =: r p'^d3I= 01z. (6) 

These equations hold true at any instant during the 
motion of the body. It will be seen, since x and ^ are the 
coordinates of the center of gravity of the body, that 
when the axis of rotation passes through the center of 
gravity, the right-hand sides of (1) and (2) become zero. 
It is further seen from (4) and (5) that if the body JS 
has a plane of symmetry as the 2:^-plane, the right-hand 
sides of these equations reduce to zero, since for every 

I xQ+ z^dM there is a corresponding I x(^ — z^dM 2iiid for 

every \ 7/Q+z)dM there is a corresponding \ y(^— z^dM. 

Therefore, ivhen the axis of rotation passes through the 
center of gravity and the plane xy is a plane of symmetry^ 
the six equations become : 

^x, = 0, , (7) 

2y, = 0, (8) 

22, = ; (9) 

2mom,-^ = 0, (1*^) 

2 mom, J,— 0, (H) 

^mom,,=^dI,. (12) 



DYNAMICS OF^MACniNERY 



185 



This case is the one that usually comes up in engineer- 
ing problems, and so these simplified equations are more 
often used than the six more general equations. It will 
be noticed that these equations are exactly the same as 
the conditions for equilibrium as determined in Art. 35 
except that 2 mom^-^ is not zero. 

105. Rotation of a Sphere. — Suppose the body B, Fig. 
125, to be a cast-iron 
sphere, radius 2 in., con- 
nected to the axis by a 
weightless arm whose 
length is 6 in. Let the 
body be rotated by a 
cord running over a 
pulley of radius 1 in. 
situated 2 in. below P^. 
Call the constant ten- 
sion in the cord 10 lb. 
and suppose it acts in 
the y^-plane. Suppose 
a=l ft. and 6 = 6 in. 
Consider the motion 
when the sphere is in 
the a:^-plane. Take 
the xy-j^lane through 
the center of the sphere 
perpendicular to z, then 

( xzdM and ( yzdM 'avq both zero. Using the foot-pound- 
second system of units, we have, x = ^ = -^, ilif = |, i^ = 
.065, (? = 8.05 1b., and 




Fig. 125 



186 APPLIED MECHANICS FOR ENGINEERS 



^- = ^'^-^^--8-?' 


(a) 


2y,= -10-P',-P,= -^-«g^ 


(^) 


23, = P,- 8.05 = 0; 


(0 


2 mom,, = -P',- 10(f) + P,(|) = 0, 


(^) 


2mom,, = P',-(8.05)(i)-P.^ = 0, 


(0 


2momi, =lf = 6'(.065). 


(/) 



From equation (/) we get ^=12.81 radians per (second)^. 
Suppose the body begins to rotate from rest and that at 
the time under consideration it has been rotating one 
second. From the relation co = 00^+ 6t (Art. 97) we get 
€0 = z= 12.81 radians per second. Solving the remaining 
equations, we get P'^= - 3.623 lb.; P'^ = -1.507 lb.; 
P,= 8.05 lb.; P,=- 15.281 lb.; P^ = 13.616 lb. The 
negative signs indicate that the arrows in the figure have 
been assumed in the wrong direction. 

Problem 157. A fly wheel 3 ft. in diameter rotates about a verti- 
cal axis. The cross section of the rim is 3 in. x 3 in. and is made of 
cast-iron. Neglect the weight of the spokes. This wheel is placed 
on the axis in the preceding problem instead of the sphere. If the 
other conditions are the same, find the reactions of the supports. 

Problem 158. The sphere in Fig. 125 is replaced by a right 
circular cast-iron cone of height one foot and diameter of base one 
foot. The vertex is placed at the point of attachment of the sphere. 
If the other conditions are the same, find the reactions of the sup- 
ports. 

Problem 159. In the problem of the sphere, Art. 105, find the 
angular velocity at the end of 30 sec. and the reactions of the sup- 
ports for such speed. 



i^* 



I -1. ^ C I 1 5^ ,-u 

DYNAMICS OF MACHINERY 



106. Center of Percussion. — It 
will be interesting now to consider 
the motion of a slender homoge- 
neous rod due to an impulsive 
force P when free to swing about 
a horizontal axis through one end. 
Let the rod be given in Fig. 126 
and let be the axis perpendicu- 
lar to the paper about which rota- 
tion takes place. The length of the 
rod is Z, and P is the impressed 
force tending to produce rotation. 

The effective forces are repre- 
sented in the figure as acting on 
each individual particle, equal in 
each case to dM • a^ = dMOp, where 
p is the distance of dM from 0, 
D'Alembert's Principle for the 
horizontal forces gives 



y- ^ r 



187 







-- -i — X Y" 

c; 
-^ > 



-f{0> — p. 



e 



Fig. 126 



P - P^= jdM0p = e^pdM= 0pM; 
similarly moments about the axis through (?, 

Pd = 0^pHM== 61,. 
But Z for a slender rod has been shown to be 



- Ml^ and /o = -, 



so that 



p^ = p-eMp = p-^Mp=p(i-^^^ 



188 



APPLIED MECHANICS FOR ENGINEERS 



H 



If Fj, = 0, then d = ^l. Under such conditions, if the 
rod were struck with a blow P, there would be no hori- 
zontal reaction at 0. This point distant 1 1 from 0, for 
which Pj, = 0, is called the center of percussion. The 
general problem of center of percussion is not quite within 
the scope of the present work. 

A right circular cylinder of height h and diameter of 
base d is made of cast iron. Locate its center of percus- 
sion, when supported as the rod in Fig. 126. 



107. 



Compound Pendulum. — When a body rotates about 
a horizontal axis due to the action of 
gravity, it is called a compound pen- 
dulum. We have seen how to find 
the time of vibration of a simple pen- 
dulum and can investigate its motion 
completely, for small oscillations. We 
shall now study the motion of the 
compound pendulum. 

Let the pendulum be represented 
by Fig. 127 and suppose the axis of 
rotation is through perpendicular 
to the paper. Taking moments about 
the axis of rotation, we have 




Fig. 127 



— Grd sin a= 6L 



0' 



e= - 



Grd sin a _ Grd sin a _ gd sin a 



L 



Mkl 



hi 



It is seen that varies with sin a. We wish now to 
find the length of a simple pendulum that will have the 
same period of vibration as this compound pendulum. 



DYNAMICS OF MACHINERY 189 

It was found, Art. 88, thtat the tangential acceleration for 
a simple pendulum, a^ = — ^sina, and hence its angular 

acceleration = — ^ -. Equating this value to the 

c 

value of found for the compound pendulum, we get 

as the length of a simple pendulum having the same 
period of vibration. This length I is the length of the 
compound pendulum. That is, the length of a compound 
pendulum is the length of a simple pendulum having the 
same period of vibration. 

The student may find this length experimentally by 
taking a piece of thread with a small lead ball attached 
to one end, and holding the other end at adjust the 
length of the thread until the compound pendulum and 
the simple pendulum vibrate simultaneously. The length 
of the thread is the length Z. 

Since I = — ^, and k^ = k§, + (P^ 



we may write 



d(l -d} = kL or 00' ' aa^ = constant, 



Evidently the relation is not changed if 0' and 0^' be 
interchanged; we may, therefore, say that the compound 
pendulum will vibrate with the same period when sus- 
pended about 0^' as an axis. This point is called the center 
of oscillation. The point is known as the point of sus- 
pension. The result may be stated as follows : m a 



190 



APPLIED MECHANICS FOR ENGINEERS 



compound pendulum the ]joiyit of suspension and the center 
of oscillation are interchangeable. 

The time of vibration of a simple pendulum was found 

tobei^ = 7r^/- (Art. 88). This gives for the time of 
vibration of a compound pendulum, 



gd 

Problem 160. A cast-iron sphere whose radius is 6 in. vibrates 
as a pendulum about a tangent hne as an axis. Find the period 
of vibration and the length of a simple pendulum having the same 
period. Locate the center of oscillation. 

Problem 161. A steel rod one inch in diameter and 3 ft. 

long is free to turn about a horizontal axis through one end. 

AVhile hanging from this axis it is suddenly acted upon by a 10-lb. 

force perpendicular to its length in such a way as to cause the hori- 
zontal component of the 
reaction of the support to 
be zero. How far from 
the support does the force 
act? 

Problem 162. Two 

drums whose radii are ?'i = 
16 in. and r^ = 12 in. are 
mounted as shown in Fig. 
128. Their combined 
weight is 200 lb. and Jc = 
U", The forces Gi and 
6^2 3,ct upon the drum, as 
well as journal friction 
amounting to 16 lb. The 
radius of the shaft is 1 in. 
Find the velocities of Gi 
and G'2 and the drums when the point of attachment of the cord on 




20 LB& 



Fig. 128. 



DYNAMICS OF MACHINERY 191 

the small drum has traveled from rest at A to a point A '. Neglect 
the friction at B, 

108. Experimental Determination of Moment of Inertia. — 
The computation of the moment of inertia of many bodies 
is a difficult matter. It is often convenient, therefore, to 
use an experimental method in dealing witli such bodies. 
The compound pendulum furnishes a means whereby 
such determinations may be made. From Art. 107, we 
find that the time of vibration of a compound pendulum 
is 



This may be written 



gd 



^ = 5^^ ' 



multiplying both sides by M^ the mass of the body, we 
have 

It thus appears that if c?, the distance from to the 
center of gravity, is known (the center of gravity may be 
located by balancing over a knife edge) and also the 
weight Gr^ and the body be allowed to swing as a pendu- 
lum about as an axis, t may be determined, giving J^. 

If I^, be desired, it may be determined from the formula 
(see Art. 41), 

Problem 163. The connecting rod of a high-speed engine tapers 
regularly from the cross-head end to the crank-pin end. Its length is 
10 ft., its cross section at the large end 5.59 " x 12.58'' and at the 



192 APPLIED MECHANICS FOR ENGINEERS 

cross-head end 5.59'^ x 8.39'^ Neglecting the holes at the ends, the 
center of gravity is 6J: in. from the cross-head end. The rod is 
made of steel and vibrates as a pendulum about the cross-head end in 
1.3 sec. Compute its moment of inertia. 

Problem 164. The student should take such a connecting rod as 
the one in the preceding problem and by swinging it as a pendulum 
find its period of vibration. Compute the moment of inertia. 



109. Determination of g. — From the preceding article 
we see that 12 2 r *> 



this relation enables us to determine gr, as soon as we know 
Jq, i^f, and c?, by determining the time of vibration about the 

r 2 

point 0. It is evident that __o is a constant for the 

^ Md 

body, when the axis is through 0, and that when once 

determined accurately the pendulum might be used to 

determine g for any locality. 

7-2 

This constant,— , is known as the pendulum constant 

Md^ ^ 

Problem 165. A round rod of steel 6 ft. long is made to swing 
as a pendulum about an axis tangent to one end and perpendicular to 
its length. The rod is 1 in. in diameter. Determine the pendu- 
lum constant. 

Problem 166. The center of gravity of a connecting rod 5 ft. 
long is 3 ft. from the cross-head end. The rod is vibrated as a pendulum 
about the cross-head end. It is found that 50 vibrations are made in 
a minute. Find the radius of gyration with respect to the cross-head 
end. 

110. The Torsion Balance. — A torsion balance consists 
of a body such as ABQ^ Fig. 129, suspended by means 
of a slender rod or wire rigidly clamped at both ends. 
Suppose the wire clamped at 0, and let the body ABC 



DYNAMICS OF MACHINERY 



193 



/'UU/AW//////' 



I) 



be a cast-iron disk of radius r and thickness t. The point 
of support is the center of gravity. The mark OA 
on the body is shown in the neutral 
position. The application of a certain 
torque in the plane of the disk causes 
it to turn through a certain angular 
distance so that OA assumes period- 
ically the positions OB and 00^ due 
to the resistance of the wire. It is well 
known that a circular rod or wire when 
twisted offers a resistance to the twist, 
such that the resisting torque varies as 
the angle of displacement. As the line 
OA moves to OB the resisting torque 
offered by the wire steadily increases. 
After the body has been given a twist ^ig. 129 

the only forces tending to produce rotation are the 
forces in the wire. So that if we call the moment of the 
couple m we may write m = ca^ since the moment of 
resistance varies with a. If, for the particular wire in 
question, when a = a^ that m = m^ we may write, since c 

is constant, ^= -^• 

Taking moments about the axis of rotation, we have 




m = eL 



zi 



or 



la 

a. 



L 



codco 
da 



since (odco = Oda-t and the resisting torque is negative. 
Multiplying through by da and integrating, we get, if 
when a = a, 



(O 



'0' 



194 APPLIED MECHANICS FOR ENGINEERS 

or co = ^JP^^/al-^a^. 

But ft) = -— , so that 



cZ^ 



c?^ 



=^/5 



c?a 



^1 Va2_^2 



If we integrate, taking ^ = 0, when a = a^, we have 

where ^ is the time taken in turning from OB through 
any angle a. Suppose the upper limit zero (a = 0), then 



sin~i -^ = 0, TT, etc. 
Suppose sin~^ — = tt, 

then « = |V^- 

The value of t given represents one fourth of a com- 
plete backward and forward swing, so that for a complete 
period 

t = 2.^lI^. . 

It is seen that this time of vibration is independeyit of the 
initial angular displacement a^. 



DYNAMICS OF MACHINERY 



195 



It is also seen that the moment of inertia of a body 
might be determined by suspending it as the body ABC 



m 



is suspended. The constant (? = -— i is a constant of the 

wire or rod and depends upon the material and diameter. 
Knowing this constant, it would only be necessary to 
determine the period of vibration in order to find /. 
For practical purposes, however, it is desirable to 



m 



For this 



eliminate from consideration the value 

purpose suppose the disk provid.ed with 

a suspended platform rigidly attached 

as shown in cross section in Fig. 130. 

Let t be its time of vibration and I its 

moment of inertia about the axis of 

suspension. Now place on the disk two 

equal cylinders IT in such a way that 

their center of gravity is the axis of 

suspension. Let t^ be the period of 

vibration of the cylinders and support 

and 7j their moment of inertia. 

t^ I 
Then — = — The moment of inertia of the two cyl- 

ti I, 

inders with respect to the axis of rotation is known; 
Then I^ = 1+ 1^, 




Fig. 130 



call it I^, 



SO that 



J=Z 



^2 



2^2_^2 



This gives the moment of inertia of the torsion balance, 
which, of course, is a constant. 

The moment of inertia of any body L may now be 



196 APPLIED MECHANICS FOR ENGINEERS 

determined by placing the bod}^ on the suspended plat- 
form with its center of gravity in the axis of rotation and 
noting the time of vibration. Calling the time of vibra- 
tion of the body L and the balance t^ and their moment 
of inertia io, we have 

3 "^ 

Let the moment of inertia of L itself be /^, so that 

Then ^^ = ^^- 

This method may be used in finding the moment of 
inertia of non-homogeneous bodies, provided the center 
of gravity be placed in the axis of rotation. 

Problem 167. The moment of inertia of a torsion balance is 
6300 and its time of vibration 20 sec. The body L consists of a 
homogeneous cast-iron disk 3 in. in diameter and 1 in. thick. Find 
the time of vibration of the balance when L is in place. Assume 
the moment of inertia of the disk. 

Problem 168. The same balance as that used in the preceding 
problem is loaded with a body L, and the time of vibration is found 
to be 30 sec. Determine the moment of inertia of Z. 

111. Constant Angular Velocity. — The six general equa- 
tions may be written, 

PJ + P^ + '2x = OMy - oflMx, 
PJ + P^ + S^ = - BMx - ofiMy, 

PJ + P, + ^z=0, 



DYNAMICS OF MACHINERY 197 

FJa + P^b + 2(2X- xZ-) = oy^-jxzd3I+ O^yzdM, 

where the P's represent the action on the bearings and 
2X, Ey, and EZ represent the components of the forces 
producing the rotation and ^(^yZ—zY}. etc., the moments 
of these forces. Now if JT, y, and Z are each zero, the 
last equation shows that ^ = 0, and therefore the angular 
velocity co is constant. The condition, however, that A"^, 
y, and Z be each equal zero, means that the forces tend- 
ing to produce rotation no longer exist. Such an axis 
is sometimes called a permanent axis. It is a principal, 
axis of the body for the point. 

112. Rigid Body Free to Rotate. — If in addition to the 
conditions that X, Y, Z, \xzdM eind \yzdMhe each zero, 
we impose the condition that both x and y be zero, that 
is, that the 2:-axis pass through the center of gravity, the 
equations of motion become 

PJ + P. = 0, 

P>+P,5 = 0, 
^ = 0. 

The body is in equilibrium under the action of the 
reactions of tlie supports and continues to rotate with 
uniform velocity co about the original axis of rotation. 



198 APPLIED MECHANICS FOR ENGINEERS 

The axis of rotation is now a principal axis of the body 
through the center of gravity. It is often called an axis 
of free rotation. Since there are three principal axes of 
the body through the center of gravity, there are three 
free axes of rotation. 

113. Rotation of Symmetrical Bodies. — When a homo- 
geneous body having a plane of symmetry rotates with 
constant angular velocity cd about an axis perpendicular to 
that plane, the only forces acting on the body reduce to 

where P is the centripetal force acting through the 
center of gravity and p is the distance of the center of 
gravity of the body from the axis of rotation. The force 
of gravity is assumed to produce no rotation. Let the xy- 
plane be the plane of symmetry and suppose the axes to 
rotate with the body and the ^-axis be the axis of rota- 
tion. It is seen that equations (1) and (2) of Art. 104 
are now identical and each expresses the fact 

1.x = ofiMp, 

and the other four equations are satisfied by this condition. 

This will be more easily understood by applying it 
especially to the sphere in Fig. 125. The only force act- 
ing, if o) is constant and the rry-plane is the plane of 
symmetry, will be a centripetal force P=co^Mp^ if we 
neglect the tendency to rotate about the x and y axes on 
account of the weight of the sphere. 

If the body is also symmetrical with respect to the 
axis of rotation, we may consider for each half, P= oy^Mp^ 



DYNAMICS OF MACHINERY 



199 



where M is the mass of | of the 
body and p is the distance of that 
one half from the axis of rotation. 
As an illustration consider the ro- 
tation of a fly wheel. Suppose all 
the centrifugal force is carried by the 
rim and neglect the spokes. If the 
mean diameter of the wheel is r (Fig. 
131), its mass M, and it rotates with 
constant angular velocity co about its 
axis, then 

2F=ay'^Mp. 




Fig. 131 



2r 



If the rim be considered as a thin wire, p = — (Prob. 23) 

TT 

and M= ^ irrF^ so that 

9 

It is seen that the tension in the rim varies with the 
square of the angular velocity. 

In particular, suppose the wheel made of cast iron and 
let r= 6 ft. and #= 10^' x4'^ Then 

P = 0)2(140.1). 

If the speed of rotation is six revolutions per second, we 

^^^^ 0)2=1421.29 and 

P= 199,122 lb. 

Dividing by ^=40 sq. in., tlie area of cross section, we 
get the stress on tlie material in pounds per square inch 
as 4978, 




200 APPLIED MECHANICS FOR ENGINEERS 

Problem 169. If the wheel just described should "run wild," 
what speed would be attained before the bursting of the rim occurred, 
supposing the rim to carry all the centrifugal forces? Assume the 
tensile strength of cast iron as 25,000 lb. per square inch. 

114. Rotation of a Locomotive 
Drive Wheel. — The drive wheel of 
a locomotive, Fig. 132, may be con- 
sidered for the present as rotating 
about a fixed axis. We shall con- 
sider the effect of the weight of the 
counterbalance on the tire due to ro- 
tation only, on the assumption that 
the tire carries all the weight of the 
l^ 132 counterbalance. 

Note. It is to be understood that the wheel center carries part 
of the weight of the counterbalance, but a complete solution of the 
problem of the drive wheel is beyond the scope of this book. The 
above assumption is therefore made. 

Let M be the mass of ^ of tire and p the distance of its 
center of gravity from the center of wheel. Let M^ be 
the mass of the counterbalance, and p-^ the distance of its 
center of gravity from the center of wheel. 

Then 2 P = 0)2 (Mp + M^p^) . 

In particular suppose the diameter of the tread of the tire 
to be 80 in. ; distance of the center of gravity of ^ of tire 
from center 27 in., and mass of | of tire 21. The mass 
of the counterbalance is 20, and the distance of its center 
of gravity from the center of the wheel 29 in. Substitut- 
ing these values, we get 

2 P =0)2 [ 21(f J) + 20 (fl)] = 95.5o)2. 



DYNAMICS OF MACHINERY 



201 



If now we know the speed of rotation of the wheel so 
that CO is known, we may determine P. Let us take co 
corresponding to a speed of train of 60 mi. per hour. 
This gives co = 2QA radians per second and 

F = 33,380 lb. 

Problem 170. If the area of a cross section of tire is 20 sq. in. 
under the assumption given above, the stress on the metal due to 
rotation about the axis would be P divided by 20, or 1669 lb. per 
square inch. 

Problem 171. If the allowable stress on the metal is 20,000 
lb. per square inch, the value of o) necessary to develop such a 
stress is given by 



0) 



= M 



20,000 X 20 
47.7 



= 91.5 radians per second. 



This corresponds to a speed of train of 207 m. per hour. 

Problem 172. Consider the rotation to take place about a point 
on the track. Find P for a speed of train of 60 rni. per hour. 
Find the corresponding stress in pounds per square inch. What 
speed of train would develop a stress in the tire of 20,000 lb. per 
square inch? 



115. Rotation about an Axis not a Gravity Axis, 
the center of gravity of a 
rotating part of a machine 
is not on the axis of rota- 
tion, there is a force tend- 
ing to bend the shaft equal 
to ofiMp. The distance 
p is the distance of the 
center of gravity from 
the axis. 

Suppose the body be a 
disk of steel, Ficy. 133, Fig. 133 



•When 




202 



APPLIED MECHANICS FOR ENGINEERS 



radius 6 in., and tliickness 1 in., and let to == 60 tt 
radians per second. If the disk is off center -| in., the 
force perpendicuhir to the sliaft due to the unbalanced 
mass is given by the equation 

= (60 7r)Va)2 J,- II 0^ (^,) =1241 lb. 

Such a disk might be balanced by the addition of a 
proper Aveight placed with its center of gravity diametri- 
cally opposite the center of gravity of the disk and in the 
plane of the disk. 

For static balancing it would not be necessary for the 
added weight to have its center of gravity in the plane of 
the disk, but for rotation this is necessary, as will be 
shown in what follows. Let the shaft AB^ Fig. 134, 

. . carry weie^hts G- 



']' 



¥L 



G, 




Fig. 134 



and (7^, as shown. 
When Gr is up, it 
tends to lift the 
end A of the shaft, 
due to its centrif- 
ugal force. When 
Gti is up, the end 
B of the shaft is lifted. Thus for each revolution, the 
end ^ is lifted, and then the end B ; that is, the shaft 
wobbles about the point (7. What really happens is 
something more than the mere lifting of the shaft in 
the bearing. Each end describes a cone with as 
the apex. 

One wheel, improperly balanced, when rotated on a 
shaft, causes the shaft to wobble about one of the bearings 



DYNAMICS OF MACHINERY 203 

or to lift bodily. Two wheels improperly balanced on 
the same shaft, cause the shaft to wobble about some 
point C. 

The principle involved in the balancing of rotating 
parts is made clear by considering two masses, M^ and M^, 
distant r-^ and r^-, respectively, from the axis of rotation. 
Suppose these bodies to be in the same plane. For 
static balance it is necessary that M^r^ = M^r^, but for 
dynamic balance, in addition to this, we must have 
M-^r^ = M^r^. It follows, therefore, that for static balance 
an equivalence of moment is required^ while for dynamic 
balance an equivalence of both masses and distances is re- 
quired. 

If the counterbalance on the locomotive drive wheel 
(see Fig. 87) does not balance perfectly the parts on the 
opposite side of the center and the reciprocating parts, 
the unbalanced mass will cause the locomotive to lift up 
when it is above the center. When it is below the center, 
the weight comes down on the rail with what is known as 
a "hammer blow." This becomes very destructive to 
both rail and wheel at high speeds when the unbalanced 
mass is at all large. It is much the same in effect as the 
dropping of the weight on the driver through a given 
distance. 

Suppose the counterbalance is too heavy by 64.4 lb., 
and that its center of gravity is 30 in. from the center of 
the wheel. If the locomotive is making 60 mi. per 
hour, and the drivers are 80 in. in diameter, the approx- 
imate lifting force when the counterbalance is up is, 
from P= (omp, 3780 lb. For a speed of 100 mi. per 
hour, this lifting force would be about 10,980 lb. In 



204 



APPLIED MECHANICS FOR ENGINEERS 



either case this weight applied suddenly to the rail must 
be very destructive to both wheel and rail. 

Problem 173. A steel disk 3 ft. in diameter and 1 in. thick 
is not perpendicular to the axis of rotation, but is out of true by 
yJo of its radius. Find the twisting couple introduced tending to 
make the shaft wobble. 



Problem 174. If the unbalanced weight in a drive wheel in the 
above illustration is 200 lb., find the centrifugal force for a speed of 
train of 60 mi. per hour. 

116. Rotation of the Fly Wheel of Steam Engine. — Let 
the fly w^heel be given as in Fig. 135 w4th radius r, and 
suppose that a belt runs over it horizontally as shown by 




— -. n ^- 



-2a- 



FiG. 135 



Pj and P^. The effective steam pressure is P; iV^ is the 
pressure of the guides on the crosshead. (It is normal if 
friction is neglected.) The pressure on the crank pin is 
resolved into tangential and radial components 2^ and Ny 
The relation between P^ and P^ is shown by the expres- 
sion Pj =: (const.) P2 = C'A (se^ ^i^t. 156). The six 
general equations give, considering only the fly wheel, 



DYNAMICS OF MACHINERY 205 

^x = -I\- I\^ - N^ cos a - Tsin a = 0, 

^y = Tcosa — iVj sin a — (? = 0, 

2^ = 0, 
Smom^= 0, 
2momj,= 0, 
2 mom^ = F^r - P^r + Ta= 61,. 

It is reasonable to assume that the resistance of the 
machinery, as shown by P^ and P^^ is constant. The last 
equation, then, states that jT, the tangential force on the 
crank pin, varies with ^, the angular acceleration. From 
this equation we have, remembering that P^ = CT^g' C' < 1 

or P^ > Pj, 

Ta-(P,^P,)r ^^^ 



L 



Since P^ > Pi, it is evident that the numerator will be 
zero when Ta = QP^ — P-^r^ so that 6 will be zero for such 
a case. This makes the angular velocity to either a maxi- 
mum or a minimum at such a point. At the dead point 
B^ T is zero; as a increases, 2^ increases until at a certain 

point B^ it equals (^P^ — P^-, At this point ^ = and 

o) is a minimum. Beyond Pj, o) increases and 6 increases, 
T has a maximum value and so does 6 at some point beyond 
B^^ after which ^ decreases, since it is again zero at the dead 
point A. Thus in passing to zero there is a value such 

that T= (P^ — P\)~'> so that is again zero at some point 

Ay At this point Ay, 6 changes from positive to negative, 
so that 0) is a maximum. It is evident that there are two 
corresponding points A-^ and B^ below the line AB. 



206 APPLIED MECHANICS FOR ENGINEERS 

The above equation may be written 

codco = Y ) ^^^^ "" T I ^^^^ 

where the subscript zero indicates some initial value ; now 
ada = ds^ distance in the crank-pin circle and rda = c?s', 
distance in the fly-wheel circle. We may, accordingly, 
write 






= Crds - (Po - A) arc of fly wheelT- 

Since work done (see Art. 135) on one end of connecting 
rod equals the work done on the other, i Tds = I Pdx^ 

where x^ x^^ and dx are distances in the cylinder corre- 
sponding to s, Sq, and ds in the crank-pin circle. Assum- 
ing that this has already been shown, we may write 

(0)2 -«2) ^x Y 

I^- — 2~^ = j -^^^ ■" ^^2 - Pi) arc of fly wheel . 



^0 



The approximate value of I Pdx may be found by reading 

from the indicator card the values of P for successive 
values of x between the limits x and x^, 

A more exact treatment of the above equation may be 
obtained by considering that the expansion of steam in 
the cylinder is constant and equal to P' up to the point 
of cut-off and that beyond this point the pressure varies 
inversely as the volume. If we assume P constant and 
equal to P^ to the cut-off, then the limits of integration 
will be regarded accordingly, and we may write 



DYNAMICS OF MACHINERY 207 



i;^!l__^ ^ J^^f% - (P2 - i^i) arc of fly wheel 



?i 



= F'(x^-Xq) - {F^-F^) arc of fly wheel 

Beyond the cut-off F varies inversely as the volume of 
steam in the cylinder, and so F = q[ -— ) = 2i. Then 

\ ITV^X J X 

from the point of cut-off to the end of the stroke 



-.s' 



I,- —^ = ?i (A - ^1) ^^I'c of fly wheel 

If the pressure be regarded as constant throughout, 
that is, if the mean effective pressure F^' be substituted 
for P, we have, considering the motion from B to J., 

(^2 -col) 
I. ^ 2 = ^i2 a - (P, - P^vrr. 

The two first equations of this article in SX and E Y 
give ^ ^ (jr ^Qg ^ _ ^p^ ^ p^>^ gjj^ ^^ 

and 

■^ \ since / "^ 

Problem 175. Suppose the mean effective steam pressure is 16,000 
lb., the radius of the crank-pin circle 18 in., and the radius of the fly 
wheel 3 ft. If {P^ ~ ^\) — ^^^ ^h. and w^ = 2 tt radians per second, 
find iDA\ if 4=2000. 

Problem 176. The fly wheel in the above problem has a velocity 
0)^ = 6 TT radians per second. What constant resistance (P^ — P^) will 
change this to 2 tt radians per second in 100 revolutions? 

Problem 177. Find the values of a for which co and are maxi- 
mum and minimum in the above problem. 



208 



APPLIED MECHANICS FOR ENGINEERS 



117. Rotation and Translation. — In this work only the 
simple case of rotation and translation in a straight line 
will be considered, since the engineer is not usually con- 
cerned with more complicated motions. Let us consider 
the motion of a body rotating about an axis that moves 
parallel to itself. Suppose the body in Fig. 136 rotates 

about 0, and at 
the same time has 
a motion of trans- 
lation along X. 
Take the origin 
at 0, and allow it 
to be translated 
with the body. 
Any elementary 
mass dM of the 
body may be con- 
sidered subjected 
to a force dM- a 
parallel to X, due 
to the translation, a tangential force dT= 6Mp^ and a 
normal force dN = co^Mp, These are the effective forces. 
Writing down the equations of equilibrium between these 
forces and the impressed forces, we have at any instant 

2a; = J dMa+ ydTsin a- JdJ^icosa =Ma+ 0My -co^Mx 

^y = - f d]Sr sin a- fdTcos a= - ay'^My - OMx 




Fig. 136 



23=0 




2mom^ 


= 


2inomj, 


= 



2 mom^ ^ '^ J^'^P~~ J ^^^P sin cc = — 61^ - a3Iy. 



DYNAMICS OF MACHINEBY 



209 



It is seen at once that if the x and 7/ axes pass through 
the center of gravity, so that x and y are each zero, the 
right-hand side of the second equation becomes zero, and 
the first and last equations may be written 

^x = 3Ia, 
2 mom^ = — 61^, 

As an illustration let us assume that a cast-iron cylinder 
rolls on a straight horizontal track, due to the application 




Fig. 137 



of certain impressed forces. Suppose the radius of the 
cylinder is 18 in. and its thickness 2 in., and that at the 
time of observation it is making 10 revolutions per 
second. From this time there is only a constant tangen- 



210 APPLIED MECHANICS FOR ENGINEERS 

tial force of friction F acting parallel to X and the 
normal pressure iV. The cylinder comes to rest in one 
minute. Find F; the distance passed over, in coming 
to rest ; the angular velocity at the end of 10 sec, and 
the linear velocity at the end of 30 sec. Take the origin 
at the center and X horizontal. From Fig. 137 it is seen 
that the general equations for this case become 

F=:Ma, 

N= a, 

Fr=0Z. 

Since F is constant, a and are constant, so we have in 
addition to the above equations 

CO = COq + 6t^ 
9. 2 

20 ' 
az= ^0fi -{- (Oot. 

These equations are sufficient to determine the unknown 
quantities. 

Problem 178. The same cylinder given in the above illustration 
rolls, without slipping, down a rough inclined plane, inclined at an 
angle 8 to the horizontal. In addition to the forces acting as given 
above there is a component of gravity (see Fig. 138). 

Let a; be parallel to the plane, then 

2a; = F4- Gsm8 = May 
^y = N- GcosS = 0, 
Smom, = Fr = 6h, 



so that a is constant and equal to 



g sin 8 

1 + ^ 



DYNAMICS OF MACHINERY 



211 



If the cylinder has the same velocity as in the above illustration, 
find the constant force of friction and the distance passed over in 
coming to rest, 8 = 10°. 




Fig. 138 



Problem 179. A cast-iron cylinder, radius 3 in. and 6 in. long, 
rolls down a rough inclined plane, inclined at an angle of 30° with 
the horizontal. Find the acceleration dow^n the plane ; the angular 
acceleration ; the force of friction jp; and the normal pressure N. 

118. Side Rod of Locomotive. — The side rod of a loco- 
motive furnishes an interesting study of a case of com- 
bined rotation 




dN" 



P 



□cLV 




and translation. 
Assume the ve- 
locity of the loco- 
motive uniform 
so that each dM 
of the side rod 
revolves uni- 
formly in the arc 
of a circle of radius r. See Fig. 139 (referred to the 
locomotive). Writing the equation of equilibrium after 



Q 

Fig. 139 



212 APPLIED MECHANICS FOE ENGINEERS 

neglecting the thrust due to the pressure of steam on 
the piston, we have 

where P is the pressure on a crank pin due to the rotation 
alone and v is the tangential velocity of any dM relative 
to the locomotive. If v^ is the velocity of the train and r' 
the radius of the drive wheel, then 

r 
so that 



2P-a 



^/2 



Problem 180. Suppose the locomotive to have a velocity of 
90 mi. per hour, the radius of the crank-pin circle 20 in., the radius 
of the drive wheel 40 in., and the weight of the parallel rod 400 lb. 
Find the pressure on the crank pins due to the centrifugal force. 

119. The Connecting Rod. — The connecting rod of an 
engine has a circular motion at one end while the other 
end moves backward and forward in a straight line. We 
shall consider the motion relative to the engine and shall 
assume that the fly wheel is of sufficient weight to give 
the crank a motion sensibly uniform. It will be con- 
venient to regard the motion of the connecting rod as 
consisting of a rotation about the crosshead end while 
that end is moving in a straight line. 

In Fig. 140 let A be the crosshead and the center of 
the crank-pin circle. Let I be the length of the connect- 
ing rod, and r the radius of the crank-pin circle. If we 
neglect friction, the only forces acting on the connecting 



DYNAMICS OF MACHINERY 



213 



rod at A are iV^', the pressure of the guides, and P\ the 
pressure exerted by the piston rod. The force exerted on 
the connecting rod by the crank pin has been resolved 
into its normal and tangential components N^ and T^ re- 
spectively. Suppose ft)i to be the constant angular velocity 
of the crank, and suppose the angular velocity of the rod 
about A to be represented by cd and the angular accelera- 
tion by 6. 

If we consider any element of mass dM oi the rod, it is 
seen that the forces acting upon it consist of a force dP = 




Fig. 140 



dM' a parallel to OX, a normal force dN= ofiMp and a tan- 
gential force dT=6Mp, so that at any instant we have 
the same formulae as those developed in Art. 104. These 
may be written in this case. 

"Ixi = P' - T^m « - iVj cos a = Ma - co'^3Ix - OMy, 

^yi = - iV^' + iV^ sin a - ^Tcos a = - cifiMy + QMx, 
2 mom^i = N^ sin (cc + (/>) - Tl cos (« +</>)= 6L - aMy. 

The axis of rotation cannot pass through the center 
of gravity, so no further reduction is possible. 



214 APPLIED MECHANICS FOB ENGINEERS 

We know 

sin a __ Z 

sin (f) r' 
so that (f> = sin~i f - sin a j ; 

differentiating with respect to t, 

sm a J 



_ = «= 



V^2 ' 

l--sm2a 

therefore a, = ^^i'^^^'^ . 

-\/P —r^sin^a 

da 
since — -= — ft). 

dt ^ 



and 



n _ — ^\^ sin a(Z2 _ ^2^ 
(P — r^sin^ a)% 



From these last two equations, for any values of (o^ and 
a we may obtain o) and 6. The linear acceleration, a, has 
been taken the same for each dM of the rod, and equal to 
the acceleration of the piston. If the quantities M^ ^, ^, 
and I^ are known ; we might find for any steam pressure 
P' and the corresponding a, the forces N\ iVj, and T, In 
other words, we could determine the forces acting on the 
guides and crank pin. 

Problem 181. The connecting rod given in Problem 163, Art. 
108, is in use on an engine whose crank has a constant angular ve- 
locity of 26 radians per second. The length of the crank is 2 ft., 
the effective steam pressure on the piston is 16,000 lb. Let a be taken 
as 30"^, then <^ = 5° 45' ; w = — 4.52 radians per second, and = — 69 
radians per second squared. From Problem 163, / = 2172, M — 61.1, 
X = 5.3 ft., y = .533 ft. To determine a, consider the relation be- 
tween the velocity v of the piston and the velocity t\ of the crank 



DYNAMICS OF MACniNERY 



215 



pin, Fig. 140. Since the velocity of every point of the rod in the 
direction of its length is the same, the projections of v and vi on the 
rod are equal. The rela- 
tions will not be altered '''^X G 
if the figure vi\ED be 
turned through 90° and 
E be placed at and 
ri be made to coincide 
with OF and drawn to 
such a scale as to equal 
r. Then v will fall on 





^^^ 


1 "K " ^ 




'^\f^^ 


1 ^'N 









Fig. 141 



OG and will be of a length equal to that cut off on OG hj AF 
produced (see Fig. 141). 

V sin (a + d)) . , , , 

— _ V — L^r/ _ gjj^ ^ _^ (>Qg ^ ^^^-^ ^^ 

vi cos (^ 

V = vi (sin a + cos a tan <^) . 

?^p-$. = -, For small values of <fy we may replace tan </> 

sin a I 

by sin <^, so that 

V = ?;i(sin a + cos a -sin a ) = n (sin a + — sin 2 a). 



Then 



But 



The acceleration a 



= vi (cos a 4- - cos 2 a)(Di. 
But since vi = <oir, 

a = (0? r (cos ci-{-'^ cos 2 a), 



For a = 30°, v = - 30.5 ft. per second, and a = 1306 ft. per (second)2. 

The three equations in ^x, %, and 2 mom;^ now give N' = 9660 lb., 
Ni = 58,137 lb., and T = - 18,151 lb. Compounding iVi and T, we get 
the resultant pressure on the crank pin to be 60,900 lb. The 
negative signs for Ni and T indicate that the arrows were assumed in 
the wrong direction. 

Problem 182. Show that the values of a that make to a maximum 
or minimum, when the motion of the crank is assumed constant, are 
TT, 37r, etc., and o, 2 tt, etc. 



216 APPLIED MECHANICS FOR ENGINEERS 

Problem 183. Find what values of a will make a maximum or 
minimum. Locate the crosshead for these values. 

Problem 184. Find values for T, N', iVi, and the resultant pres- 
sure on the crank-pin when cc = tt and when « = 0. Use the above data. 

Problem 185. Assume a force of friction F acting on the cross- 
head, such that F = .06 N\ In the above case when a = 30°, what is 
the value of F, iV, iV^i, and T? 

Problem 186. Suppose the steam pressure zero, find T, N', Ni, 
and the resultant crank-pin pressure, if coi is the same. 

120. Body Rotating about an Axis — One Point Fixed. — 
We shall now consider the case of a body rotating about 
an axis when only one point of that axis is fixed. Con- 
sider the equations 1, 2, 3, 4, 5, and 6 (Art. 104) and recall 
that a body acted upon by any system of forces may be 
considered as being acted upon by a single force and a 
single couple (Art. 36). If one point of the axis is fixed, 
the single force will act at this point, but the effect of the 
single couple will be to move the axis of rotation about 
this point. The components of the moment of the couple 
are shown in the right-hand side of equations (4) and (5). 
If these reduce to zero, the couple vanishes and rotation con- 
tinues about the original axis of rotation even though only 
one point of that axis is fixed. But this can happen only 

when I xzdM = and also ( yzdM ^ 0. We may say, 

then, that a body rotating about an axis one point of which 
is fixed ^ when no forces are acting to produce rotation^ will 
continue to rotate about that axis with a constant angular 
velocity o). 

121. Gyroscope. — The gyroscope illustrated in Fig. 142 
consists of a metal wheel A mounted on an axis BB^ fixed 



DYNAMICS OF MACHINERY 



217 



at one point to the stand 0, The weight D serves to bal- 
ance the wheel about the point of support. The wheel 
A is very delicately mounted, so that there is little fric- 
tion. It is set in motion by means of a cord wound 
around its axle, as in the case of an ordinary top. 




Fig. 142 

When the weight D exactly balances A so that BB is 
horizontal, we have a case of a body rotating about an 
axis fixed at one point, with no external forces acting. 
According to the previous article, the body continues to 
rotate about that axis. 

If, however, the weight I) does not balance A^ so that 
BB is not horizontal, the axis of rotation changes, since 
in that case the force of gravity tends to turn BB about 
a horizontal axis through JS perpendicular to BB, As a 
result of the two rotations, the body tends to turn about 
a new axis, so that BB turns about the point B horizon- 



218 APPLIED MECHANICS FOR ENGINEERS 

tally. All this is in accordance with Art. 95, where we 
saw that the resultant of two angular velocities was an 
angular velocity given by the diagonal of a parallelogram 
constructed upon the two velocity arrows as sides. This 
rotation of the gyroscope about the vertical axis through 
JE is known, as precession. The student may reproduce 
the above results experimentally by taking a bicycle wheel 
mounted upon its axle. Suspend one end of the axle by 
a string and hold the other in the hand, so that the axle 
is horizontal. With the other hand now give the wheel 
a spin. If the axle remains horizontal, the wheel contin- 
ues to spin about the same axis, but if the hand support- 
ing one end of the axle be removed, the wheel continues 
to rotate about its own axis while the axis rotates about 
the suspending string. In other words, the wheel has a 
motion of precession. 

The motion of the bicycle wheel is explained in the 
same manner as that used in explaining the precession 
of the gyroscope. 

122. The Spinning Top. — The student will be interested 
in seeing that tlie motion of a spinning top, with which 
all are familiar, is also capable of the same explanation. 
Let the top be represented, as in Fig. 143, with its point 
at 0, and suppose it has an angular velocity co about its 
own axis. If it is rotating sufficiently rapidly, and its axis 
is vertical, it " sleeps," or continues to revolve about that 
same axis. If, however, the axis be tilted slightly from 
the vertical, the weight of the top Gr and the reaction of 
the floor constitute an unbalanced couple tending to make 
it revolve about an axis through perpendicular to the 



DYNAMICS OF MACHINERY 



219 



paper. The result of tliese two rotations is to cause tlie 

top to always tend to revolve about an axis a little in 

front of the geometrical 

axis. This causes the 

axis of spin to continually 

advance, and describe a 

cone about OZ, that is, 

the top precesses. It is 

well known that when the 

top has been so disturbed, 

if spinning rapidly, it 

moves about OZ very slowly, and gradually takes the 

vertical position again. 

When the velocity of rotation co finally becomes small, 
the irregularities of the support throw the axis out of the 
vertical, and the action of the unbalanced couple causes 
precession. At first the precession is very slow, but grad- 
ually increases as (o decreases until the top falls. 




123. Motion of Earth. — A brief presentation of this 
subject would be incomplete without mentioning the pre- 
cession of the earth's axis. In Fig. 144 let S represent the 
sun and U the earth, with its axis slightly inclined to the 
vertical. The earth is a spheroid with its axis of rotation 
as its short axis. Consider the ring of matter near the 
equator which if cut off would make the earth spherical. 
The attraction of the sun for this ring of matter is greater 
on the side nearest the sun. This causes the earth to be 
acted upon by an unbalanced force i^, and tends to cause a 
rotation of the earth about an axis through perpendicu- 
lar to the paper. As a result the axis of rotation is moved 



220 



APPLIED MECHANICS FOR ENGINEERS 



forward slightly so that its path is a cone about OZ. On 
account of the very great velocity of the earth and the 
smallness of the force F^ this precession is very slow, a 




^F 




Fig. 144 



complete rotation of the axis about OZ requiring 25,800 
years. (See Young's '' General Astronomy," Precession of 
the Equinoxes.) 

124. Plane of Rotation. — We have seen that a body 
moving in a straight line continues to move in that line 
due to its inertia unless acted upon by some force (Art. 76). 
In a similar way a body rotating about an axis tends to 
maintain its axis and plane of rotation due to its moment 
of inertia unless acted upon by some external forces. The 
moment of inertia of a rotating body has the same relation 
to its rotation as the inertia of a body has to its translation. 

The student may get some idea of this tendency of a 
rotating body to maintain its plane and axis of rotation 
from the study of a bicycle wheel mounted on its axle. If 
the wheel be held by grasping both ends of the axle and if 
it is then rotated rapidly, it will be found that there is no 
difficulty in moving the rotating wheel in the plane of ro- 



DYNAMICS OF MACHINERY 



221 



tation, but that as soon as an attempt is made to change 
the direction of the axle, there is considerable resistance. 
The same experience may be had by treating the wheel of 
the gyroscope (Art. 121) in a similar way. 

The same action takes place in the rolling of ships at 
sea. The rolling is lessened by the tendency of the large 
fly wheels on board to maintain their plane of rotation. 

125. Gyroscopic Action Explained. — Spinning bodies, 
such as the gyroscope or a fly wheel, when acted upon by 
an unbalanced couple that produces a rotation at right 
angles to the spin, rotate about an axis at right angles to 




Fig. 145 



each of these (Art. 124). Such action may be explained 
as follows (see Fig. 145). Suppose the body to be a disk, 



222 



APPLIED MECHANICS FOR ENGINEEBS 



and that rotation is taking place about the axis OY in the 
direction indicated. A couple shown by the forces P 
produces rotation about the axis OZ in the direction in- 
dicated. 

Consider any element of the disk dM; it is subjected to 
two angular velocities, co about OY^ and o)-^ about OZ. If 
we imagine the element at the point A on the axis OZ, its 
velocity v due to rotation about OZ is pco^ = 0, since 
p = 0. As it moves from A toward B this velocity v in- 
creases, and from B to it decreases until it is zero at 0. 
This increase and decrease of the velocity is represented 
in the figure by the arrows v. Since the velocity of dM 
increases in going from A to B^ the increase must be caused 
by some force having the direction dP^^ on the side facing 
the reader. The decrease in the velocity v in going from 
B to must be due to a force dJP^^ acting away from the 
side facing the reader. In a similar way the velocity v 
increases in going from O to I) and decreases in going from 
I) to J.. The force acting on any dM as it moves from 
A to A again are represented by the arrows dPy It 
is seen that the result of such forces acting on every dM 
will be to turn the disk about the axis OX. This motion 

about OZ we have called preces- 
sion (Art. 121 and Art. 122). 



126. Precessional Moment; 

Special Case. — A simple analysis 
serves to give the moment about 
the axis OX. Let the disk of the 
preceding article be represented 
in Fig. 116 with the axis OY 




Fig. 146 



DYNAMICS OF MACIHNEHY 223 

perpendicular to the paper. The elementary mass dM is 
in the position shown. The velocity (y^ of this dM'i^ 

V = po)^ = rco^ sin «, 

when p is the distance from the axis (9Z, co^ the angular 
velocity about OZ^ and r is the distance from 0, 

Therefore the acceleration a-^ in the direction of v is 

given by 

dv da 

— = rco. cos a — . 

dt ^ dt 

But — = ft), the angular velocitv about OY^ 
dt 

so that a^ = rco^co cos a 

and cZPj, the force in the direction of v, 
= dM' a^ = dJfrco^co cos a. 

We may write dM= t±rda dp. 

Then dP. = t'^(o.(or\Ir cos ada. 

9 

The moment of this force about the axis OX is 

mom^^ = dP^(r cos «) = t-co^cor^dr cos^ «(?«. 

The moment for the whole disk U about the axis OX is 
then 



JJ= t-o).co I r^dr i cos^ ada 

y r^ n^ ^ -, .y r^ 
= c-ft).ft)— I cos^ ada = t~o).co~ 

9 ^^0 9 ^ 

= t-CO.CO—TT, 

9 ' ^ 



- + - Sin 2 a 
Z 4 



22-i APPLIED MECHANICS FOR EJSIGINEEBS 

But M=tl7rr\ 

9 

so that U= JJwiw^ . 

If the rotating body is the rim of a fly wheel with out- 
side radius r^ and inside r^^ the value of U changes only 
in integrating between the limits r^ and r^. Then 

TT /7 W-^D ivr (^i+i) 

As an illustration, suppose the weight of the ring of a 
gyroscope to be 50 lb. and the mean radius 6 in., and 
outside radius 8 in., and let co^ be unity, and suppose it 
makes 100 revolutions per second about the axis OY. 
Then 

It is seen that with a small value for co^ the tendency 
to turn about OX is considerable. 

Note. The above analysis is substantially the same as that 
given in Engineering, June 7, 1907. 

Problem 187. A ship carries a cast-iron fly wheel whose rim is 
6 ft. outside diameter, 4 in. thick, and 18 in. wide. When it 
is making 3 revolutions per second, its axis is turned about an axis 
through the plane of the wheel with unit angular velocity. Find the 
moment of the couple that tends to turn the wheel about an axis 
perpendicular to these two axes. 

Problem 188. A solid cast-iron disk 3 ft. in diameter and 3 
in. thick revolves about its axis, making 3000 revolutions per 
minute. At the same time it is made to turn about an axis in its 
plane at the rate of 2 revolutions per minute. Find the magnitude 
of the couple tending to rotate the disk about an axis perpendicular 
to these two axes. 



DYNAMICS OF MAC U INERT 



225 



The equation 
may be written 



U = M(i).u}— 
4 



(i). 



O) 



If the value of the couple U is constant, it is seen that o varies in- 
versely with oiy That is, if o) is large (o^ is small, and if oj is small coj 
is large. This was pointed out in Art. 122 and illustrated in the case 
of the top when it is dying down. As the spin decreases in such 
cases, the precession increases. 

127. Precessional Moment ; General Case. — The preces- 
sional moment for any body when OZ is perpendicular to 
01^ may be obtained from a consideration of the moment 
equation (Art. 126) by retaining the dM. The equation 
may then be written 

U— \ dP^Qr cos «) = j dMo)^(or^ cos^ « 



or 



U= (o^cD I (iil!f(r cos a)2. 



Since r cos a is 
the distance of 
dM from OX, 
this may be i' 
written 






When the axis 
of precession OZ 
is inclined at an 
angle S to the 
axis of spin OY, 
the above value 
for JJ must be 

Q 




Fig. 147 



226 APPLIED MECHANICS FOR ENGINEERS 

changed slightly. To make the ideas clear, let the orig- 
inal axis OZ be drawn as well as the inclined axis OZ' 
making the angle 8 with OY^ as in Fig. 147. Let p be 
the distance of an elementary dMivom OZ, as before, and 
Pj the distance of dM from OZ^, Then p = pi sin 8, and 
the expression for the velocity, v = po)^ = rco^ sin a becomes 

V = rcwj sin a sin S. 

and a.= -— = "^^i^ sin B cos a, 

at 

since sin S is constant, so that 

dP^ = ai^dM= dMrco^co sin S cos a ; 

therefore U= j dP-^{r cos a)= j dMcoiCo sin Sr^cos^^ 

= cD^o) sin S I c?illf (r cos a)^, 
or Z7= Wjwjr^^sinS. 

The gyroscopic action of the fly wheel of an automobile 
has much to do in causing the machine to overturn when 
rounding sharp curves at high speeds. Even when the 
machine is not overturned, the gyroscopic moment due to 
the rotation of the wheel causes an extra pressure on the 
bearings. This pressure is shown by the wear on the 
bearings. It is left as an exercise to determine the over- 
turning moment due to the action of an automobile fly 
wheel under assumed conditions. It will also interest the 
student to know that a German torpedo boat 116 ft. long, 
and 56 tons displacement, was held upright in a heavy 
sea by an 1100 lb. disk rotating 1600 revolutions per 
minute. 



DYNAMICS OF MACHINERY 227 

Problem 189. A locomotive is going at the rate of 40 mi. per 
hour around a curve of GOO ft. radius. The diameter of the drivers 
is 80 in., and a pair of drivers and axle have a moment of inertia 
about an axis midway between the wheels and perpendicular to the 
axle of 3000. What is the magnitude of the couple introduced by 
the precessional motion of this pair of wheels? Give tlie direction in 
which it acts. Does it tend to make the locomotive tip inward or 
outward ? 

Problem 190. A car pulled by the locomotive in the preceding 
problem has four pairs of wheels. The moment of inertia of each 
pair of wheels and their connecting axle, with respect to an axis mid- 
way between the wheels and perpendicular to the axle, is 320 (see 
problem 87). What is the magnitude of the precessional couple acting 
upon the whole car ? 

Problem 191. The fly wheel of an engine on board a ship makes 
300 revolutions per minute. The rim has the following dimensions : 
outside radius 4 ft., inside radius 3J ft., width 12 in. The ship 
rolls with an angular velocity of ^ a radian per second; find the torque 
acting on the ship due to the gyrostatic action of the fly wheel. 

Problem 192. A conical top is made of wood and is spinning 
about its axis with a velocity of 20 revolutions per second. The cone 
has a base of 2 in. and a height of 2 in., and spins on the apex. 
While spinning steadily with its axis vertical (sleeping), it is dis- 
turbed by a blow so that its axis is inclined at an angle of 30° with 
the vertical. Find the velocity of precession and the torque U that 
tends to keep the top from falling. See Fig. 143. 

128. Car on Single Rail. — An interesting application of 
the gyroscope has been made recently in England. A 
car (see Fig. 148) is run upon a single rail, and is 
held upright by means of rapidly rotating fly Avlieels. 
Each car contains two of tliese wheels rotating in op- 
posite directions, at the rate of 8000 revolutions per 
minute. 



228 



APPLIED MECHANICS FOR ENGINEERS 



Any tendency of the car to tip over, either when run- 
ning or standing at the station, is righted by the gyro- 
scopic action of the fly wheels. The experimental car 





O' 






"0 






.^.^Lbrx 






r^ 1 1^-. 




i;cy \^j) 


tcJL—Lu) 


'^ ^/i i^A M ^A (^A f^At^iyAtd <^J i^i (^A e^ i/Ai^ (yA l^-J CA t<) iA ^A t>> t^ kA ^' 






Fig. 


148 







was so successful in operation that it maintained itself in 
an upright position even when loaded eccentrically. The 
action of the fly wheels is such as to place the center of 
gravity of the car and load directly over the rails. 



(Note. See Engineering^ June 7, 1907.) 



CHAPTER XIII 



WORK AND ENERGY 



129. Definitions. — When the forces acting upon a body 
cause a motion of that body, work is done. We define 
the work done by a force as the magnitude of the force 
times the distance through which the hody^ upon which it acts^ 
moves along its line of action. 

This definition may be less exactly stated by saying 
that a force acting on a body that moves through a dis- 
tance does work. This brings to mind the forces consid- 
ered as acting in Chapters II, III, and VI, where no mo- 
tion was produced; that is, where the point of application 
did not move. Such forces produce lio work according 
to our definition. ■ 

To make the idea 1^ 

of work clearer, 
suppose the body 
(Fig. 149) to 
be acted upon by 
a force P, and 
that the body is moved until the point at A is finally 
at B, The work done by P is P times AB. Suppose 
the plane upon which C^ moves is rough, so that it offers 
a resistance F. In passing over the distance AB, the 
force F does a work of resistance equal to F times AB. 

229 




Fig. 149 



230 APPLIED MECHANICS FOP ENGINEERS 

The upward force iV, which is the pressure of the support- 
ing surface, does no work, since no motion takes phice along 
its line of action. The idea of work is related to that of 
the motion of the body in the direction of the acting force, 
but is independent of time. 

130. Units of Work. — Since work involves force times 
distance, we express it in terms of the units of force and 
distance; that is, in inch-pounds or foot-pounds. These are 
the units used by engineers in this country and England, 
and are the units that will be used in this book. 

We might say, then, that the unit of work, the foot-pound^ 
is the work done hy a force whose magnitude is one pound 
when the body upon which it acts moves through a distance 
of one foot. 

In countries where the metric system is used, the erg 
is used when a small unit of w^ork is convenient. The 
erg is the work done by a force of one dyne when the body 
upon which the force acts moves a distance of one centi- 
meter in the direction of the force. A larger unit of 
work, the joule, is often used; the joule is 10" ergs. En- 
gineers often use the kilogram-meter as a unit of work. 
It is the work done by a force whose magnitude is one 
kilogram, while the body upon which the force acts moves 
one meter in the direction of the force. 

131. Graphical Representation of Work. — Work has been 
defined as the product of a force and a distance. If the 
force be uniform and equal to P, and the body upon 
which it acts be moved through a distance a, the graph- 
ical representation of the work done by P is given by the 



WOBK AND ENERGY 



231 



airea of a rectangle, Fig. 150, constructed on F and a as 
sides, since P 



W= Fa, 



Fig. 150 



If the force F varies as the dis- 
tance through which the body is 
moved along its line of action, we 
may represent the work by the area of the triangle as 
shown in Fig. 151. Let the force be zero when the 

motion begins, and let it be F^ 
when the distance passed over along 
its line of action is OA. Then since 
the force varies as the distance, it 
is equal to F for any intermediate 
distance 00. The total work done, 
then, in moving the body through a 
distance OA by the variable force 
P, which varies as the distance, is 

equal to — J^^- L It is seen that 

p 

this is the same as the work done by the average force —^ 

acting through the distance OA. The resistance of a 
helical spring varies with the elongation or compression. 
The same law of variation holds for all elastic bodies. 

Another variation of force with distance with which 
the engineer is frequently concerned, is the case where 
the force varies inversely as the distance through which the 
body is moved. If F is the force and S the distance, 
the relation between force and distance may be expressed, 

F = ^^H^, or FS= const. But this represents the equi- 




FiG. 151 



232 



APPLIED MECHANICS FOR ENGINEERS 



J 


^ 


s 


p^^ 1 


E 


C 


1 


D 




1 


1 







Fig. 152 



lateral hyperbola. This will be made clearer by reference 
to the specific example of the expansion of steam in a steam 

cylinder. (See 
Fig. 152.) Up to 
the point of cut- 
off 0^ the steam 
pressure is the 
same as that in 
the boiler (prac- 
tically), and is 
constant while 
the piston moves 
from to C. 
At this point, the entering steam is cut off and the work 
done must be done by the expansion of the steam now in 
the cylinder. According to Mariotte's Law, the pressure 
varies inversely with the volume of steam; but since the 
cross section of the cylinder is constant, we may say that 
the pressure varies inversely as the distance. From the 
properties of the curve, it is easy to see that the area under 
the curve represents 
the work done. 

The curve ob- 
tained in practice 
representing the re- 
lation between the 
force and distance is 
shown in Fig. 153. 

The curve after 
cut-off is not a true hyperbola, and its area is determined 
by means of a planimeter or by Simpson's Rule. 




Fig. 153 



WOBK AND ENERGY 233 

132. Power. — The idea of work is independent of time. 
But for economical reasons it is necessary to take into 
consideration this element of time. We must know 
whether certain work has been done in an hour or ten 
hours. For such information a unit of the rate at which 
work is done has been adopted. This unit is called power. 
Power is the rate of doing work. It is the ratio of the work 
done to the time spent in doing that work. 

The unit of power is the horse poiver. This has been 
taken as 550 ft. -lb. per second, or 33,000 ft. -lb. per minute. 
Originally the idea of the rate of work was connected 
with the rate at which a good draft horse could do work. 
This value as used by Watt was 550 ft. -lb. per second. 
The horse power of a steam engine is mean effective 
pressure times distance traveled by the piston per second, 
divided by 550. 

133. Energy. — Energy is the capacity for doing work; 
it is stored-up work. Bodies that are capable of doing 
work due to their position are said to possess potential 
energy. Bodies that are capable of doing work due to 
their motion are said to possess kinetic energy, A familiar 
example of potential energy is the energy possessed by a 
brick as it is in position on the top of a chimney. If the 
brick should fall, its energy at any instant would be 
called kinetic. When the brick strikes the ground, work 
is done in deforming the ground and brick, or perhaps 
even breaking the brick and even generating heat. The 
work done by the brick when it strikes is sufficient to 
use up all the energy that the brick had when it 
struck. 



23J: APPLIED MECHANICS FOR ENGINEERS 

134. Conservation of Energy. — The kinetic energy of 
the brick spoken of in the last article was used up in 
doing work on the ground and air, and upon the brick 
itself, so that the kinetic energy that the brick possessed 
when it struck was used up. It was not, however, de- 
stroyed, but was transferred to other bodies, or into heat. 
Such transference is in accord with the well-known prin- 
ciple of the conservation of energy. This principle may 
be stated as follows : energy cannot he created or destroyed. 
The amount of energy in the universe is constant. This 
means that the energy given up by one body or system of 
bodies is transferred to some other body or bodies. It 
may be that the energy changes its form into light, heat, 
or electrical energy. 

Energy cannot be created or destroyed; it is, therefore, 
evident that such a thing as perpetual motion is impossible. 
Such a motion would involve the getting of just a little 
more energy from a system of bodies than was put into 
them. i./i / 

135. Energy of a Body moving in a Straight Line. — Sup- 
pose the body, Fig. 154, moving in a straight line as in- 



^^-^ 



n 



>^ 



n 



Fig. 154 



dicated with a variable velocity v. Let P be a constant 
working force, so that the resulting force acting on the 
body will be P — R. From the relation that the accelerat- 



WORK AND ENERGY 235 

ing force equals the mass times the acceleration, we may 

= if~ 

Integrating between the limits v^ and t', and and s, 

we have — -^=(P-R}s, 

The quantity --^ is the kinetic energy of the body 

when it has a velocity v^ and the quantity -^ the kinetic 

energy of the body when it has a velocity v^. The left- 
hand side of the equation, therefore, represents the change 
in kinetic energy. The equation shows that tlie work 
done hy the tvorhing force equals the ivorh done hy the re- 
sisting force plus the change in the kinetic energy. 

The weight of the body is 64.4 lb., and P is a constant 
force, say 100 lb., and R a constant resistance = 84 lb. 
If the body starts from rest, what will be the velocity when 
it has moved a distance of 16 ft.? 

Substituting in the above equation, w^e have z; = 16 ft. 
per second. 

Problem 193. A car whose weight is 20 tons, and having a veloc- 
ity of 60 mi. per hour, is brought to rest by means of brake friction 
after the power has been shut off. If the tangential force of friction 
of 200 lb. acts on each of the 8 wheels, how far will the car go 
before coming to rest? 

Solution. Here M = -^ — ^ v^ — 88 ft. per second, i; = 0, P = 0, 



236 APPLIED 3IECHANICS FOR ENGINEERS 

since there are no working forces, it = 1600 lb., so that 

40,000 (88)2 = 1600 5. 
2(32.2) ^ ^ 

Therefore s =3009 ft. 

Problem 194. Suppose the car in the preceding problem to be 
moving at the rate of 60 mi. per hour when the power is shut ofP, 
what tangential force on each of the 8 wheels will bring the car to 
rest in one half a mile ? 

Problem 195. What is the kinetic energy of a river 200 ft. wide 
and 15 ft. deep, if it flows at the rate of 4 mi. per hour, the weight 
of a cubic foot of water being 62.5 lb. ? What horse power might be 
developed by nsing all the water in the river ? 

Problem 196. The flow of water in Niagara River is approxi- 
mately 270,000 cu. ft. per second. What is its kinetic energy? 
What horse power could be developed by using all the water ? 

Problem 197. This amount of water, 270,000 cu. ft., goes over the 
falls of Niagara every second. The height of the falls including rapids 
above and below is 216 ft. What horse power could be developed by 
using all the water? What horse power could be developed by using 
the water, considering the height of the falls to be 165 ft., the height 
of free fall ? 

Note. It is estimated that the total horse power of Niagara Falls, 
considering the fall as 216 ft., is 7,500,000. The Niagara Falls Power 
Company diverts a part of the volume of water above the rapids into 
their power plants, where it passes through a tunnel into the river 
below the falls. The turbines are 140 ft. below the water level, and 
each one is acted upon by a column of water 7 ft. in diameter. The 
estimated power utilized in this way is 220,000 horse power. The 
student should estimate the horse power of each turbine, assuming 
the w^ater to fall from rest through 140 ft. For a full account of the 
power at Niagara Falls, the student is referred to Proc. Inst. C. E., 
Vol. CXXIV, p. 223. 

When the motion of the body is not along the line of 
the force, as is the case in Fig. 155, where the body is 




WOBK AND ENERGY 237 

supposed moving up the plane under the system of forces 
shown, we resolve the force into components along and 
perpendicular to the direc- 
tion of motion. It is evi- 
dent that the component 
perpendicular to the direc- 
tion of motion can do no 
work in moving the body ^ ' Fig~155 

up or down the plane. 

The same might be said of the component of Gr perpen- 
dicular to the plane and of iV, so that the work-energy 
equation in such a case includes only the components of 
the forces along the line of motion. The accelerating 
force in this case is P cos a, and the resisting forces are F 
and Gr sin a. The work-energy equation becomes 

^-.^^^Fs + Ca sin a)8= (P cos «)5, 

2 2 ^ 

where s is a distance measured along the plane. 

Problem 198. A body whose weight is 32.2 lb. is pulled up an 
inclined plane, inclined at an angle of 30^ w^ith the horizontal, by a 
horizontal force of 250 lb. The motion is resisted by a constant 
force of friction of 10 lb. acting along the plane. If it starts from 
rest, what will be its velocity after it has gone up a distance of 100 ft. ? 

Problem 199. The same body as that in the preceding problem is 
projected down the plane with a velocity of 5 ft. per second. How far 
will it go before coming to rest? 

Hint. In this case there is no working force acting, and the final 
kinetic energy is zero, so that the work-energy equation reduces to the 
expression : the work of resistance equals the initial kinetic energy. 

Problem 200. The student should solve Problem 112 by using 
the principle of work and energy. 



238 APPLIED MECHANICS FOR ENGINEERS 

136. Work under the Action of a Variable Force. — When 
the forces are not constant, the work-energy equation 
already derived does not hold true. In such a case the 
equation vdv = ads becomes, by integrating, 

2 2 ^0 ^0 

The integrals expressed cannot be determined until it is 
known how H and P vary. In any case, however, we see 
that the quantity under the integral sign represents work, 
and so we may say: the work done equals the resistance 
overcome plus the change in kinetic energy. 

Let us suppose that the resistance H varies as the dis- 
tance, and also that the force P varies as the distance. 
Then R = const, x (6-) = C-^s and P = const, x (s) = O^s. 
The work-energy equation becomes, upon substitution, 






In Art. 81 the case of a body of 644-lb. weight falling 
in a resisting medium was discussed. We may now dis- 
cuss this same problem by means of the principle of work 
and energy. In this case, 



E = 10s,F= a,M= 20, v^ = V2 (/h = 62. 1 ft. per second. 

Then 20.^-2^62^ 

2 2 ^0 ^0 

^2^ 3864 --§2+ 64.4 s. 

2 

This gives a relation between velocity and distance. The 
student should complete the problem, as outlined in Art. 81. 



WORK AND ENERGY 



239 



Problem 201. A body whose weight is G4.4 lb. falls freely from 
rest from a height of 5 ft. upon a 200-lb. helical spring. Find the 
compression in the spring. 

It will be recalled from Problem 113 that a 200-lb. spring is such a 
spring as would be compressed 1 in. by a weight of 200 lb. resting 
upon it. It will also be recalled that in compressing such a spring, 
the resistance of the spring is at first zero, and that it increases in pro- 
portion to the compression. So in the present case we may write 

-rr- = — , where the distance of one inch is expressed in feet : since 
1 s 

"12" 

s is expressed in feet and R is the resistance of the spring in pounds, 
R = 2400 s. In this case P= G,v = 0, vq = ^2 gh = ^"^-^ ^^' P^^ second, 
and AI= 2. Then the work-energy equation gives 

- ^ (^^'^y ^ 2400 — = 64.4 5, 

2 2 

.s=.544ft., or 6.52 in. 

Problem 202. A weight of 500 lb. is to fall freely from rest 
through a distance of 6 ft. The kinetic energy is to be absorbed 
by a helical spring. Specifications require that the spring shall not be 
compressed more than 2 in. Find the strength of the spring required. 

Problem 203. It requires 2000 lb. to press a certain sized nail 
into a board a distance of 2 in. The same size nail is to be driven 
to the same depth by a 5-lb. hammer (Fig. 156) in 4 blows. With 
what velocity must the 
hammer strike the nail 
each time? Assume that 
all the energy of the ham- 
mer is absorbed by the 
nail and that the resist- 
ance offered by the timber 
in question varies as the 
distance of penetration of 
the nail. Neglect the 
weight of the hammer as a working force. Under the assumptions 
made, the penetration of the nail will be the same for each blow. 



3 



S~? 




Fig. 156 



\' 



V'^^VOJ 



240 



APPLIED MECHANICS FOE ENGINEERS 




Fig. 157 



Problem 204. Specifications state that it shall require 32,000 lb. 
to compress a helical spring 1^ in. What weight falling freely 
from rest through a height of 10 ft. will compress it one inch? 

Problem 205. The draft rigging of a freight car shown in Fig. 

157 is provided with 
two helical springs, one 
inside the other. The 
outside spring is a 10,000- 
Ib. spring, and the inside 
a 5000-lb. spring. A car 
weighing 60,000 lb. is 
provided with such a 
draft rigging. While go- 
ing at the rate of 1 mi. per hour it collides with a bumping post. 
How much will the springs be compressed ? 

Problem 206. The draft rigging in the preceding problem is 
attached to the first car of a freight train, consisting of 30 cars, each 
weighing 60,000 lb. How much will the springs of the first car be 
elongated if there is 10 lb. pull for each ton of weight when the speed 
is 40 mi. per hour? The speed is increased to 45 mi. per hour. 
How much will the springs be elongated if the resistance per ton at 
this speed is 12 lb. ? 

Problem 207. The Mallet compound locomotive (Railway Age, 
Aug. 9, 1907) is capable of exerting a draw-bar pull of 94,800 lb. 
According to the preceding problem, how many 60,000-lb. cars can be 
pulled at 45 mi. per hour? What strength of spring would be 
necessary for the first car, if the allowable compression is 11 in.? 

Problem 208. An automobile going at the speed of 30 mi. per 
hour comes to the foot of a hill. The power is then shut off and the 
machine allowed to "coast" up hill. If the slope of the hill is 1 ft. 
in 50, how far up the hill will it go, if friction acting down the plane 
is .06 Gy where G is the weight of the machine ? 



137. Pile Driver. — A pile driver consists essentially of 
a hammer of weight Gr so mounted that it may have a free 



WORE AND ENERGY 



241 



. 



fall from rest upon the pile (Fig. 158). The safe load 
to be placed upon a pile after it has been 
driven is the problem that interests the 
engineer. This is usually determined by- 
driving the pile until it sinks only a certain 
fraction of an inch under each blow, then 
the safe load is a fraction of the resistance 
offered by the earth to these last blows. 
This resistance is very small when the pile 
begins to penetrate the earth, but increases 
as penetration proceeds, until finally 
due to the last blows it is nearly con- 
stant. If we regard the hammer Gr 
as a freely falling body, and consider 
the hammer and pile as rigid bodies, 
and further assume, as is usually done, 
that It for the last few blows is constant, 
we may write the work-energy equation. 



Fig. 158 



^Mv 



1= r Rds= Rs^, 



since the final kinetic energy is zero and the weight of 
the hammer as a working force is negligible. The 
distance s^ is the amount of penetration of the pile for 
the blow in question. But v'^=2 gh^ so that 



lMvl=Gh. 



We have then as the value for the supporting power 
of a pile, 



11 = 



Gh 



2-i2 APPLIED MECHANICS FOR ENGIXEEBS 

A safe value, H^ from l to |- of i2, is taken as the safe 
supporting power of piles. The factor of safety and the 
value of s-^ for the last bloAv are usually matters of specifi- 
cation in any particular work. This is the formula given 
by Weisbach and Molesworth. Other authorities give 
formulae as follows : 

Trautwine, Ii=QO G -Vh^ if s^ is small, 

, ^5 aVh 

and U = — ; 

^1 + 1 

Wellington, H = — , where h is in feet and s-^ in inches; 

McAlpine, i? = 80 [ (7 + (. 228 VA - 1)2240] ; 
Goodrich, jB = -— -. 

For other formulae and a general discussion of the 
subject of the bearing power of piles, the reader is referred 
to Transactions of Am. Soc. C. Eng., Vol. 48, p. 180. 

The great number of formulae for the supporting power 
of piles is due to the various assumptions that are made 
in deriving them. In deriving the Molesworth formula, 
the hammer and pile were considered as rigid bodies. 
It will be seen that the hammer and pile are both elastic 
bodies, both are compressed by the blow; there is friction 
of the hammer with the guides, and the cable attached 
to the hammer runs back over a hoisting drum. There 
is, in most cases, a loss of energy due to brooming of the 
head of the pile. This broomed portion must be cut off 
before noting the penetration due to the last few blows. 

Results of tests have also been taken into consideration 



WORK AND ENERGY 243 

and have modified the formula}. The Wellington formula 
differs from the Molesworth formula only in the denomi- 
nator, where s^ + 1 is used instead of s^. This has been 
done to guard against the very large values of R given 
when §1 is very small. According to the Wellington 
formula, M can never be greater than (7A, the total energy 
of the hammer, and this is perhaps the safer formula to 
use for small values of s^. 

Engineers have come to believe that it will be extremely 
difficult to get a general formula that will give very ex- 
act information as to the bearing power of piles, since 
soil conditions are so varied. The more simple formulae 
with a proper factor of safety are used. 

As an illustration of the use of the formula, let us 
consider the problem of providing a pile to support 75 
tons. If the weight of the hammer is 3000 lb., and the 
height of fall 15 ft., the pile will be considered down when 

Gh 3000 X 15 o p^ o r • 
s. = -— - = = .o it. = o.b m. 

^ E 1^0,000 

Using a factor of safety of i, we have s^ = .6 in. 

Problem 209. Compute the value of s^ for the pile in the above 
illustration by using the various formulae given in this article. 
Compare the results. 

Problem 210. A pile is driven by a 4000-lh. hammer falling 
freely 20 ft. What will be the safe load that the pile will carry if 
at the last blow the amouut of penetration was | in.? Use a factor 
of safety of J. Compute by the Molesworth and the Wellington 
formulae, and compare. 

Problem 211. A pile was driven by a steam hammer. The last 
twenty blows showed a penetration of one inch. If two blows of 
the steam hammer cause the same penetration as one blow from 



244 



APPLIED MECHANICS FOB ENGINEERS 



-2r- 



B 



r 



a 2000-lb. hammer falling 20 ft., what weight in tons will the pile 
support? Assume the penetration for each of the last few blows 
the same. 

138. Steam Hammer. — The steam hammer consists 
essentially of a steam cylinder mounted vertically and 

having a weight or hammer attached 
to one end of the piston rod. Let 
AB (Fig. 159) be the steam cylinder, 
D the piston, and jETthe anvil, upon 
which a piece of metal is shown un- 
der the hammer (7. The steam pres- 
sure in the cylinder is constant and 
equal to P, while the piston passes 
over a distance a to cut-off, and 
varies inversely as the volume over 
the remaining distance h, (r, the 
weight of the hammer and piston, is 
also a working force. The resistance, 
I d -B^ of the metal varies 

I during any blow with 

amount of compression. It is 



^55 



Q 



t 



K 



the 



Fig. 159 



zero just as the hammer touches the 
metal, and increases up to a maxi- 
mum when the compression is great- 
est. Let R^ be the average resistance of the metal, and 
iZj, the exhaust pressure. Then the work-energy equa- 
tion for the hammer before striking the metal becomes 

Ml.M^R^ p. = P p, + Cp'as + a Cds, 



or 



2 2 



+ E^s = Pa+ Gs +j^P'ds. 



WORK AND ENERGY 245 

Since P' varies inversely as the volume of the cylinder, 

., jy, const. 
we may write, I^' = — =-• 

Then the work-energy equation gives 

^ + E.s = Pa+ Gs + c\ogt. 
2 a 

The term — ^ ^^ zero, since the motion has been consid- 
ered from rest at the top of the cylinder to a distance s. 
The quantity c may be computed by reading from the 
indicator card the value of P' at s. It will be seen that 
8 has been taken greater than a ; that is, the piston is 
beyond the point of cut-off. 

When the hammer finally comes to the face of the 
metal, the work-energy equation may be written 

^ + R^h^ = Pa+ ah^ + c^ log -, 

where the distance h^ represents the value of s when 
the hammer just touches the metal, and v^^ 6', and c^ are 
the corresponding values of ^;, 5, and c. This equation 
gives the kinetic energy of the hammer when it strikes 
the metal. The work-energy equation for the hammer 
during the compression of the piece may now be written 

where d is the amount of compression of the metal due 
to the blow. This is shown by the small figure to the 
right, where the piece of metal has been drawn to a 
somewhat larger scale. 



246 APPLIED MECHANICS FOR ENGINEEBS 

After the hammer strikes the metal, the steam pressure 
and the weight of the hammer as working forces, and the 
exhaust pressure as a resisting force, have been neglected. 
The work done by these pressures is small, since the dis- 
tance d is small. Approximately, then, the work done on 
the metal equals the kinetic energy at the time of first 
contact. 

Instead of using the value P^ = -, and computing the 

s 

integral j P'ds as indicated in the formulae, values of P^ 

and s might be read from the indicator diagram (see 
Art. 131) and added by means of Simpson's formula 
(see Art. 26). 

As an illustration of the foregoing, let us suppose the 
steam cylinder 25 in. long and 14 in. in diameter; the 
steam pressure P = 18,000 lb.; the exhaust pressure 
i?i = 2300 lb.; a=7.2 in.; d = l in.; (^=644 lb.; 
c^ = 10,800 lb. ; A' = 24 in. Substituting in the work- 
energy equation, we have for the kinetic energy of the 
hammer at the time of striking the iron, 

:^ = 11,475 ft. -lb. 

This gives a value for v^ = 33.8 ft. per second as compared 
with 11.3 ft. per second for the same weight freely falling 
through the same distance. 

Investigating now the resistance of the metal, we have, 
under the assumption already made, 

i2^c^= 11,475 ft.-lb., 
so that E' = 550,800 lb. 



WORK AND ENERGY 247 

In the above discussion we have neglected the compres- 
sion of the anvil and hammer due to the blow, and also 
the friction of the piston. 

Problem 212. Find the kinetic energy of the hammer when 
k' = 18 in. Find also v and R', using the same value of d. 

Problem 213. A steam hammer exactly similar to the one given 
in the illustration above is used with the same steam pressure. It is 
only necessary, however, for the work for which it is intended, that 
the kinetic energy of the hammer for a stroke of 2 ft. be 6000 ft.-lb. 
What weight of hammer should be used? 

Problem 214. Compute the kinetic energy and velocity of the 
hammer in the illustration (G = 644 lb.) when the piston has moved 
the full length of the cylinder (h' = 25 in.). Assume that there is 
nothing on the anvil. 

Problem 215. "What value of h' in the above problem would give 
the hammer the same velocity as it would have if it fell freely from 
rest through the height h ? Compute the kinetic energy for this 
velocity. 

Problem 216. In the illustration given above, what would be 
the value of R' if the steam pressure and G be included as working 
forces, and R^ as a resisting force, during the compression of the 
piece ? 

Problem 217. In the illustration given above, suppose that, in 
addition to the compression of the piece J in., the anvil is compressed 
.02 in. Find the value of R\ 

139. Energy of Rotation about Fixed Axis. — In Art. 103, 

where the subject of the rotation of a rigid body about 

a fixed axis was discussed, the following equation was 

derived : 

^CP\d,+P^^d^+ P'^d^ + etc.} = 61 

where the P's represent forces tending to rotate or retard 
the rotation of a rigid body about a fixed axis, the 6?'s, 



248 APPLIED MECHANICS FOR ENGINEERS 

the distances of the lines of action of these forces from 0, 
6 the angular acceleration, and I the moment of inertia of 
the body with respect to the axis of rotation. 
This equation may be put in the form 

^ ^ 2 (F\d, + P^^, + P^A + etc.) 

Now let us suppose the moment P\di is made up of 
a working moment and a resisting moment, such that 
P\d^ = P\d^-R\d^\,P\d^=P^^^d^-R'^d"^, etc. Re- 
membering that (odco= 0da^\ we may write, after clearing 
of fractions, 

I(od(o = 'E(P'\d^da" + P^^^d^da^^ + P^^^d^da^^ + etc. 

- ^(iR'\d^\dci'^+ R^\d!\du!^ + R^\d^\dci!^ + etc.). 

Let the angles which P'\, P^\'^ P^-) ^tc, make with the 
r2;-axis be called d\^ a^^^, a^^g, etc. Then d^da^' = dsi^ d^daf^ 
= ds^, d^dod^ = ds^^ and d^\da'^ = ds^\^ etc. Then if con- 
tinuity exists so that we may integrate, we may write 

I jl)da) =fp^\ds^ +Jp'^^ds^ + etc. 

^CE'\ds^\^ CB'^ds'^^^eiG., 

o)= Cp'\ds^+ r>Vs2 + etc. 

- CR^\ds^\- fp^^^ds^'^^ etc. 
Since i 0)2/= 1 (ifljdMp^ = Ji dM(copy = ^\ dMv\ 



ft)^— ft). 



WOBK AND ENERGY 



249 



the kinetic energy of the body, where co is the angular 
velocity, the left-hand side represents the difference in the 
kinetic energy of rotation of the body when its initial 
velocity is co^ and its final velocity (o. On the right-hand 
side, Fds represents work, since ds is measured along the 
line of action of P in each case. A similar statement 
could be made for the E's, so that the right-hand side 
represents the work of the working forces minus the work 
of the resisting forces. Here, then, as in simple transla- 
tion, between any two positions of a rigid body, the work 
done ly the working forces equals the work done by the resist- 
ing forces plus the change in kinetic 
energy. 

As an illustration, let us consider 
the case of two weights (see Fig. 
160), (?i=20 lb. and a^-=10 lb., 
suspended from drums rigidly at- 
tached to each other and of radii 3 
ft. and 2 ft. respectively. Let the 
weight of the two drums and shaft 
be 644 lb., and the radius of gyration 
2 ft. The radius of the axle is one 
inch and the axle friction 30 lb. The 
friction acts tangentially to the axle. 

Assume that the initial velocity co^ is one radian per sec- 
ond, and the final velocity 18 radians per second, how 
many revolutions will the drums make ? 

The work-energy equation gives 

1 (18)2 (80) - \ (1)2 (80) + (^2 2 irrn + 30 ^^/i 

= G-^2 Trr^n, 




Fig. 160 



250 



APPLIED MECHANICS FOR ENGINEERS 




where r^ is the radius of the large drum, r that of the 
small drum, r^ that of the axle, and n the number of revolu- 
tions. Making the substitutions, n becomes 

7^ = 54.8 revolutions. 

Problem 218. In the above illustration, what is the velocity of 
_ Gi and G2 when co has its 

initial and final values ? In 
what time do the drums 
make the 54.8 revolutions? 

Problem 219. The drum 
in Fig. 161 is solid and has 
a radius r and a thickness h. 
Initially, it is rotating, mak- 
ing 0)0 radians per second, 
but it is brought to rest by 
the action of a brake. The 
brake is applied from below by a force P acting at the end of the beam. 

The force of friction between the drum and brake is — , where P' is 

4 

the normal pressure exerted by the beam on the drum. The radius 
of the axle is ri, and the axle friction (.05) P'', where P'' is the pres- 
sure of the axle on the bearing (neglecting the lifting caused by P)^ 
Required the work-energy equation. 

Since the drum comes to rest, the final kinetic energy is zero, so 
that 

-- coo^/ + ^' 2 Trrn + (.05) P'' 2 Trnn = 0. 

There are no working forces, so we find the equation reducing to the 
form: the initial kinetic energy equals the work of resistance. The 
normal pressure exerted by the beam on the drum may be found by 
taking moments about the hinge of the beam. Then 



Fig. 161 



P' 



a + b 



P. 



WORK AXD ENERGY 251 

The number of revolutions turned through in coming to rest is 
designated by n. The equation then becomes 

1 ^ nrn (a + b) P ^ . ^^^ p„ o ^,^„. 

2 2 /; 

Problem 220. Suppose the drum in the preceding problem to 
be 3 ft. in diameter, 1^ in. thick, and made of cast iron. It is mak- 
ing 4 revolutions per second when the force P = 100 lb. is applied to 
the beam. The length of the drum is 6 ft., and the rim weighs twice 
as much as the spokes and hub. If ^ = 1.25 ft., a -h b = S ft., and 
r =1 in., find the number of revolutions ni that the drum will make 
before coming to rest. Assume the friction of the brake on the drum 
to be I the normal pressure, and the friction of the axle (.05) P". 

Problem 221. The drum in the preceding problem is making 
3 revolutions per second ; what force will be required to bring it 
to rest in 100 revolutions ? 

Problem 222. If the brake in Problem 220 is above instead of 
below the drum, how will the results in Problems 220 and 221 be 
changed? 

Problem 223. A square prism as shown in Fig. 162 is mounted 
so as to rotate due to the weight G. The elastic cord runs over the 
pulley B and meets the square 
at P'. The mechanism is such 
that motion begins when P is 
in the position shown, and 
ceases when the prism has made 
a quarter turn; that is, when P 
reaches P'. The diameter of 
the journal is 2 in., and the 
weight on the same is 600 lb. 
The force of friction on the 
journals is 60 lb., and on the pulley at B 10 lb. Find the tension in 
the cord when P reaches P'. The cord is elastic, and is made of such 
material that it elongates, due to a pull of 100 lb., .02 in. in each inch 
of length. What is the elongation per inch due to the fall of G as 
stated ? 




B 



Q 



Q^ 100 LB8. 



Fig. 162 



252 



APPLIED MECHANICS FOB ENGINEERS 



140. Brake Shoe Testing Machine. — The brake shoe test- 
ing machine owned by the Master Car Builders' Asso- 
ciation has been established at Purdue University. It 
consists of a heavy fly wheel attached to the same axle 
as the car wheel. These are connected with the 
engine, and may be given any desired rotation. When 
this has been obtained, they may be disconnected and 
allowed to rotate. The dimensions and weight of the 
parts are known so that the kinetic energy of the fly 
wheel and rotating parts may be computed by noting the 




Fig. 163 

angular velocity. When the desired velocity has been 
attained, the brake shoe is brought down on the car wheel. 
The required normal pressure on the shoe at A (see Fig. 
163) is obtained by applying suitable weights at B, The 
system of levers is such that one pound at B gives a nor- 
mal pressure of 24 lb. on the brake shoe. The weight of 
the levers themselves gives a normal pressure of 1233 lb. 
Provision is also made for measuring the tangential pull 
of the brake friction ; this, however, is not shown in the 
figure. 



WORK AND ENERGY 253 

The weight of the fly wheel, car wheel, and shaft and 
all rotating parts is 12,600 lb., and the radius of gyra- 
tion is V2.I6. The weight of 12,600 lb. is supposed to 
be the greatest weight that any bearing in passenger or 
freight service will be called upon to carry. The diame- 
ter of the fly wheel is 48 in., its thickness 30 in., diame- 
ter of shaft 7 in., and the diameter of the car wheel is 
33 in. The brake-shoe friction is ^ the normal pressure 
of the brake shoe on the wheel, and the journal friction 
may be assumed as (.002) of the pressure of the axle on 
the bearing. The work-energy equation for the rotating 
parts after being disconnected from the engine becomes 



2\32.2 J 2V32.2 y o^^ ^ ^4 24 



n 



+ (1233 + 12,600 + 24 a)(.002)2 tt ^ « = 0, 

since there are no working forces. 

Problem 224. The speed is such as to correspond to a speed of 
train of a mile a minute when brakes are applied. What must be 
the weight G so that a stop may be made in a thousand feet? What 
is the corresponding normal pressure on the brake shoe? 

Problem 225. If the speed corresponds to the speed of a train 
of 100 mi. per hour, what weight G would be necessary to reduce 
the speed to 60 mi. per hour in one mile? What is the normal pres- 
sure on the brake shoe necessary? 

Problem 226. If the velocity corresponds to a train velocity of 
60 mi. per hour, and the apparatus is brought to rest in 220 revolu- 
tions, the weight G is 100 lb. Find the tangential force of friction 
acting on the face of the wheel. What relation does this bear to the 
normal brake-shoe pressure ? 

Note. In the preceding problems, the ratio (the coefficient of 
friction, see Art. 146) has been taken as J. One of the important 



254 



APPLIED MECHANICS FOB ENGINEERS 



uses of this testing machine is to determine the coefficient of friction 
for different types of brake shoes. Experiment shows that it varies 
generally from ^ to i, sometimes going as high as y^. 

141. Work of Combined Rotation and Translation. — The 

relation between work and energy of simple translation 
and the work and energy of rotation about a fixed axis 
have been discussed. We shall now determine the rela- 
tion for combined rotation and translation when the axis 
remains parallel to itself. Let the body of mass M 




Fig. 164 

(Fig. 164) be in rotation with angular velocity co about an 
axis at 0, and at the same time let this axis move parallel 
to itself with a linear velocity v. At any instant the 
elementary mass dM has a linear velocity of translation 
v-^ and a tangential velocity v = cop. Its resultant velocity 
is expressed as the diagonal of a parallelogram con- 
structed upon the two velocity arrows as sides, so that 



WORK AND ENERGY 



255 



^12 -_ -y2 _|_ ^2 _(_ 2 VV^ COS (f>. 



dM 



Multiplying both sides of tliis equation by — --, we have 

J -2- = J -2- + J -Y^ + j 2—.., cos<^, 

or J -^— = J — wy + J -^+J ciitfwpvicosc^; 

but /J cos (f>= t/, and | c?il!i?/ = My=^0, since OX is a gravity 
line. At any instant w and v-^^ are constant. Therefore, 

At any instant, then, the kinetic energy of combined rota- 
tion and translation is equal to the kinetic energy of trans- 
lation of the center of gravity plus the kinetic energy of 
rotation. 

As an illustration, consider a disk of radius r and thick- 
ness h rolling without slip- 
ping down an inclined 
plane, inclined at an angle 
a with the horizontal (see 
Fig. 165). There is a 
working force (? sin a and 
a resisting force F=^ (.06) 
Gr cos a. Now the kinetic 
energy of the disk is made 
up of the sum of its kinetic energy rotation and transla- 
tion. If ft)Q and v^ be the respective initial angular and 
linear velocities, we have 

lcu2j- l,,2r+ 1 3j^2 -\mvI ■\-Fs=G sin a • s. 




Fig. 165 



256 APPLIED MECHANICS FOR ENGINEERS 

Problem 227. Suppose the disk in the above illustration to be 
made of cast iron, and let r = 2 ft., 7^ = J ft., and a = 30°. At a cer- 
tain instant it is making 2 revolutions per second. What will be 
the linear and angular velocities after the disk has gone 20 ft.? 
Would the disk finally come to rest? 

142. Kinetic Energy of Rolling Bodies. — It is convenient 
to express the kinetic energy of combined rotation and 
translation of such bodies as rolling wheels in a different 
form from that given in the preceding article. There is 
some mass M^ that will have the same kinetic energy when 
translated with a velocity v-^ as the kinetic energy of trans- 
lation plus the kinetic energy of rotation of the body of 
mass M; that is, 

2 ""22' 

for a wheel rolling on a straight track cor = Vj, where r is 
the radius. 

Then M. = M+ 4/ 

This has been called the equivalent mass. 

Applying this to the disk in the preceding article, we 
find the work-energy equation to be, 

^ ^ ^ + Fs= Grsina -s 

A ''A 

for the disk, since J= \Mr\ 31^ = f 7l!f. 

Problem 228. A sphere of radius r rolls without slipping down 
an inclined plane, inclined at an angle a to the horizontal, with an 
initial velocity r^. Show that its kinetic energy is the same as that 
of a sphere whose mass is | larger translated with a velocity Vy 



work: axd EyEBGY 257 

Problem 229. The sphere in the preceding problem is made of 
steel, 12 in. in diameter, and a = :^0^. If r^ = 10 ft. per second, wliat 
will be the velocity 10 ft. down the plane ? There is a force of fric- 
tion acting up the plane = (.03) times the normal pressure of the 
sphere on the plane. 

143. Work-Energy Relation for Any Motion. — The rela- 
tion between work and energy for the motions considered 
in this chapter holds for more complicated motions and 
for motions in general. The limits of the present work 
will not admit the proof of the general theorem. It may 
be said, however, that for any motion the w^ork done by 
the working forces equals the work done by the resisting 
forces plus the change in kinetic energy. In the case of 
the motion of a complicated machine, the total work done 
equals the total resistance overcome plus the change in 
kinetic energy of the various parts of the machine. 

144. Work done when Motion is "Uniform. — AVhen the 
motion is uniform, the change in kinetic energy is zero, 
and the work-energy equation reduces to the form : tvork 
done equals the resistance overcome. 

As an illustration, let us consider the case of a loco- 
motive moving at uniform speed and represented in Fig. 
166. Suppose P the mean effective steam pressure (see 
Art. 131), F the friction of the piston, F^ the friction of 
the crosshead, F'^ the journal friction, F^^' the crank-pin 
friction, T the friction on the rail. It the draw bar resist- 
ance, G the weight of the locomotive, and N^ and iV the 
normal reactions of tlie rails on the wheels. Consider 
only one side of the locomotive and write the work-energy 
equation for a distance s, equal to a half turn of the driver 



258 



APPLIED MECHANICS FOR ENGINEERS 



(from dead center A to dead center ^), that the loco- 
motive travels. This becomes, for the frame, 

Fira = F(7ra + 2 r) + i^'(7ra + 2 r) + E^ira - Rira, 

where R^ is the pressure of the driver axle on the frame. 
If we neglect friction, this becomes 

JP=J? -JK. 

Considering the rotating and oscillating parts, we obtain 

P(7ra+ 2r)=F(7ra + 2r) + F^ (ira + 2 r) + Tira 

+ F'^irr^ + F'^^irr^ + B^ira, 

where r^ and r^ are the radii of the driver axle and the 
crank pin, respectively. If we neglect friction, this equa- 
tion reduces to the form, 

PQira + 2r) = Tira + Wira. 




Fig. 166 

Taking all the parts represented in Fig. 166, we may 
disregard the work of friction of the cylinder and cross- 
head, since the sum would be zero. That is, the work done 
by the piston on the cylinder, due to friction, equals the 
work done by the cylinder on the piston due to friction. 
We have, then, for the work-energy equation. 



WOliK AND ENEUGY 



259 



P(ira + 2 r) = Prra + Rira + Tira + F'^ttt^ + F'^'irr^. 

It is seen that the pressure of the steam on the head of 
the cylinder, for the half of the stroke considered, is a 
resistance. If Ave neglect friction and assume perfect roll- 
ing, this equation becomes 

P Qjra + 2 r) = Pira + Bira, 



or 



2r 



or considering both cylinders, 

P = 



ira 
•1 r 



R. 



This is the formula usually given for the tractive power 
of a locomotive having single expansion engines. This 
may be expressed in 
terms of the dimen- 
sions of the cylin- 
ders and the unit 
steam pressure. Let 
p be the unit steam 
pressure in pounds 
per square inch, I 
the length of the 
cylinder in inches, d 
the diameter of the 
cylinder in inches, 
and d-^ the diameters of the drivers in inches; then 

■" d^ ' 
Considering the forces acting on the driver, disregarding 




()()(• 



EH 



Fig. 167 



260 APPLIED MECHANICS FOR ENGINEERS 

friction, and taking moments about the center of the wheel 
(see Fig. 167), we have, for uniform motions, 

Pr = Ta, 

or T=-P, 

a 

Taking moments about the point of contact of the wheel 
and rail, we have, for the position shown, 

P(a + r) = R'a, 
and since P = R — R^ 

we have R= —P. 

a 

It follows that T = R\ that is, the train resistance can- 
not be greater than the adhesion of the drivers to the rails. 

This adhesion in American practice is usually taken as \ 
to \ the load on the drivers. 

Problem 230. What resistance R may be overcome by a locomo- 
tive moving at uniform speed, diameter of drivers 62 in., cylinders 
16 X 24 in., and a steam pressure on the piston of 160 lb. per square 
inch ? What should be the weight of the locomotive on the drivers ? 

Problem 231. If the diameter of the drivers of a locomotive is 
68 in., and the size of the cylinder is 20 x 21 in., what train resistance 
may be overcome by a steam pressure of 160 lb. per square inch ? 

Problem 232. A locomotive has a weight of 155 tons on the 
drivers, if the adhesion is taken as J, this allows 31 tons for the draw- 
bar pull. The train resistance per ton of 2000 lb., for a speed of 60 
mi. per hour, is 20 lb. Find the weight of the train that can be 
pulled by the locomotive at the speed of 60 mi. per hour. 

Problem 233. An 80-car freight train is to be pulled by a single 
expansion locomotive at the rate of 30 mi. per hour. The weight of 
each car is 60,000 lb., and the resistance for this speed is 10 lb. per 
ton. What must be the weight on the drivers, if the adhesion is J? 






CHAPTER XIV 



FRICTION 



145. Friction. — When one body is made to slide over 
another, there is considerable resistance offered because of 
the roughness of the two bodies. A book drawn across 
the top of a table is resisted by the roughness of the two 
bodies. The rough parts of the book sink into the rough 
parts of the table so tliat when one of the bodies tends to 
move over the other, the projections interfere and tend to 
stop the motion. The bearings of machines are made 
very smooth, and usually we do not think of such surfaces 
as having projections. Nevertheless they are not perfectly 
smooth, and when one surface is rubbed over the other, re- 
sistance must be overcome. This resisting force to the 
motion of one body over another is known as friction. 
When the bodies are at rest relative to each otlier, the 
friction is known as the friction of rest^ or static friction. 
When they are in motion with respect to each otlier, tlie 
friction is known as the friction of motion^ or kinetic friction. 



146. Coefficient of Fric- 
tion. — If the body repre- 
sented in P'ig. 1G8 be 
pulled along tlie horizon- 
tal plane by the force P, 
the following forces will 




2G1 



262 



APPLIED MECHANICS FOR ENGINEERS 



be acting on it ; the downward force Gr (not shown) and 
the reaction li inclined back of the vertical through the 
angle 0. The reaction li of the plane on the body may be re- 
solved into two components, one horizontal and one vertical. 
The horizontal force is known as the force of friction, and 
the normal force, the normal pressure. The tangent of the 

XT 

angle ^, or — , is called the coefficient of friction. This 

coefficient of friction, which we shall represent by/, may 
be defined as the ratio of the force of friction to the normal 
pressure ; it is an absolute number. 

The coefficient of friction is usually determined by al- 
lowing a body to slide down an inclined plane as shown in 

Fig. 169. The angle 6 is 
increased until the force of 
friction F will just keep the 
body from sliding down the 
plane. The angle is then 
called the angle of repose^ and 
the tangent of is the coeffi- 
cient of friction. 

It is possible with such an 
apparatus to determine the 
coefficient of friction for various materials. It has been 
found that after motion begins the friction is less, that is, 
the friction of motion is less than the friction of rest. This 
is an important law for engineers. 




Fig. 169 



147. Laws of Friction for Dry Surfaces. — Very little was 
known of the laws of friction until within the last seventy- 
five years. About 1820 experiments were made that 



FRICTION 



203 



seemed to show that, for such materials as wood, metals, etc., 
friction varies with tlie pressure, and is independent of 
the extent of the rubbing surfaces, the time of contact, 
and the velocity. A little later (1831) Morin published 
the following three laws as a result of liis experiments on 
friction : 

(1) The friction between two bodies is directly propor- 
tional to the pressure; that is^ the coefficient of friction is 
constant for all pressures. 

(2) The coefficient and amount of friction for any given 
pressure is independent of the area of contact. 

(3) The coefficient of friction is independent of the velocity., 
although static friction is greater than kinetic friction. 

These laws of Morin hold approximately for dry unlu- 
bricated surfaces, although it has been found that an in- 
crease in speed lowers the coefficient of friction. The 
coefficient of friction is a little greater for light pressures 
upon large areas than for great pressures on small areas. 

The following is a table of some of the coefficients of 
friction as determined by Morin : 



Coefficients of Friction, due to Morin 



Material 


Condition of Surface 


e 5 

E F 


S5 






O U. Si 


III 


Brick on limestone 


Dry 


.67 


3r)° .30' 


Cast iron on cast iron 


Slightly greased 


.16 


9° 6' 


Cast iron on oak 


Wet 


.^h 


30° 2' 


Copper on oak 




.17 


9° 38' 


Copper on oak 


Greased 


.11 


6° 17' 


Leather on cast iron 




.28 


15° 39' 



2Gi 



APPLIED MECHANICS FOR ENGINEEBS 



Coefficients of Friction, due to Morin — Continued 



Material 


Condition of Surface 




1 1 










Leather on cast iron 


Wet 


.38 


20° 49' 


Leather on cast iron 


Oiled 


.12 


6° 51' 


Leather on oak 


Fibers parallel 


.74 


36° 30' 


Leather on oak 


Fibers crossed 


.47 


25° 11' 


Oak on oak 


Fibers parallel, dry 


.62 


31° 48' 


Oak on oak 


Fibers crossed, dry 


.54 


28° 22' 


Oak on oak 


Fibers parallel, soaped 


.44 


23° 45' 


Oak on oak 


Fibers crossed, wet 


.71 


35° 23' 


Oak on oak 


Fibers end to side, dry 


.43 


23° 16' 


Oak on oak 


Fibers parallel, greased 


.07 


4° 6' 


Oak on oak 


Heavily loaded, greased 


.15 


8° 45' 


Oak on pine 


Fibers parallel 


.67 


33° 50' 


Oak on limestone 


Fibers on end 


.63 


32° 15' 


Oak on hemp cord 


Fibers parallel 


.80 


38° 40' 


Pine on pine 


Fibers parallel 


.56 


29° 15' 


Pine on oak 


Fibers parallel 


.53 


27° 56' 


Wrought iron on oak 


Wet 


.62 


31° 48' 


Wrought iron on oak 




.65 


33° 2' 


Wrought iron on wrought iron 




.28 


15° 39' 


Wrought iron on cast iron 


_ 


.19 


10° 46' 


Wrought iron on limestone 




.49 


26° 7' 


Wood on metal 


Greased 


.10 


6° 0' 


Wood on smooth stone 


Dry 


.58 


30° 7' 


AVood on smooth earth 


Dry 


.33 


18° 16' 



148. Friction of Lubricated Surfaces. — The laws of fric- 
tion as given by Morin and stated in the preceding article 
hold approximately for rubbing surfaces, when the sur- 
faces are dry or nearly so ; that is, for poorly lubricated 
surfaces. If, however, the surfaces are well lubricated so 



FRICTION 2G5 

that tlie projections of one do not fit into the otlier, but 
are kept apart by a fihn or hiyer of the lubricant, tlie laws 
of Morin are not even approximately true. The study of 
the friction of lubricated surfaces, then, may be divided 
into two parts: (1) the study of poorly lubricated bear- 
ings, and (2) the study of well-lubricated bearings, the 
friction of which varies from | to -^^ that of dry or poorly 
lubricated bearings. 

Since the friction of poorly lubricated bearings is about 
the same as that of dry surfaces, we shall consider that 
the laws of oNIorin hold, and shall confine our attention to 
the friction of well-lubricated bearings. If tlie lubricant 
is an oil, the friction of the bearing is no longer due to 
one surface rubbing over the other, but to the friction 
between the bearing and the oil, and to tlie internal fric- 
tion of the oil. That is, the oil adheres to the two sur- 
faces and its own particles attract each other, and the 
motion of one of the surfaces with respect to the other 
changes the positions of the oil particles. It is to be 
expected then that the friction of an oiled bearing will 
depend upon the viscosity of the oil, upon the thickness of 
the layer interposed hetiveen the surfaces, and upon the 
velocity and form of the bearing. 

The coefficient of friction is no longer constant, but 
varies with the temperature, velocity, and pressure. The 
variation of the coefficient of friction of a paraffine oil 
with temperature is shown in Fig. 170 when the pressure on 
the bearing is 33 lb. per square inch and a velocity of rub- 
bing of 296 ft. per minute. It is seen that the coefficient 
of friction decreases with increase of temperature until a 
temperature of 80° F. is reached, when it increases rapidly. 



266 



APPLIED MECHANICS FOR ENGINEERS 



This means that above this temperature the oil is so thin 
that it is squeezed out of the bearing, and the conditions 
of dry bearing are approached. The temperature at 
which oils show an increasing coefficient of friction is dif- 




.02 .03 

COEFFICIENT OF FRICTION 

Fig. 170 



ferent for different oils, even at the same pressure and 
velocity. The curve in Fig. 170, however, may be re- 
garded as typical of all oils when the pressure and velocity 
are constant. 

The following table, due to Professor Thurston, shows 



FRICTION 



267 



the relation between tlie coefficient of friction and tem- 
perature for a sperm oil in steel bearings when the veloc- 
ity of rubbing is 30 ft. per minute. 



Pressure, lb. 


Temfekatiue. 


Coefficient 


PKESSirUE, LH. 


TeMI'KUATUKE, 


Coefficient 


PER SQ. IN. 


Degrees F. 


OF Friction 


PER SQ. IN. 


Degrees F. 


OF Friction 


200 


150 


.0500 


100 


110 


.0025 


200 


140 


.0250 


50 


110 


.0035 


200 


130 


.0160 


4 


110 


.0500 


200 


120 


.0110 


200 


90 


.0040 


200 


110 


.0100 


150 


90 


.0025 


200 


100 


.0075 


100 


90 


.0025 


200 


95 


.0060 


50 


90 


.0035 


200 


90 


.0056 


4 


90 


.0400 


150 


110 


.0035 









It is seen that for a pressure of 200 lb. per square inch as 
the temperature increases from 90° F. the coefficient in- 
creases, indicating that the temperature of 90°, for the 
given pressure and velocity, was above the temperature at 
which the oil became so thin as to be squeezed out and the 
bearing to approach the condition of a dry bearing. For 
a constant temperature 110° F. and 90° F. the coefficient is 
seen to decrease with increase of pressure up to a certain 
point and then to increase. This is a typical behavior of 
oils when the temperature is constant and the pressure 
varies. 

At speeds exceeding 100 ft. per minute, the same 
authority found '' that the heating of the bearings within 
the above range of temperatures decreases the resistance 
due to friction, rapidly at first and then slowly, and 
gradually a temperature is reached at which increase 



268 



APPLIED MECHANICS FOR ENGINEERS 



takes place and progresses at a rapidly accelerating 
rate." 

The relation between the coefficients of rest and of 
motion as determined by Professor Thurston for three oils 
is given below. The journals were cast iron, in steel 
boxes; velocity of rubbing 150 ft. per minute and a tem- 
perature 115° F. 





g 


PERM Oil 


West 


Virginia Oil 




Lard 




Pressure, 

LB. PER. 
SQ. IN. 


At 150 

ft. per 

min. 


At 

start- 
ing 


At 
stop- 
ping 


At 150 
ft. per 
min. 


At 
start- 
ing 


At 
stop- 
ping- 


At 150 
ft. per 
min. 


At 
start- 
ing 


At 

stop- 
ping 


50 


.013 


.07 


.03 


.0213 


.11 


.025 


.02 


.07 


.01 


100 


.008 


.135 


.025 


.015 


.135 


.025 


.0137 


.11 


.0225 


250 


,005 


.14 


.04 


.009 


.14 


.026 


.0085 


.11 


.016 


500 


.004 


.15 


.03 


.00515 


.15 


.018 


.00525 


.10 


.016 


750 


.0043 


.185 


.03 


.005 


.185 


.0147 


.0066 


.12 


.020 


1000 


.009 


.18 


.03 


.010 


.18 


.017 


.0125 


.12 


.019 






Steel Journals an 


d Brass Boxes. 








500 


.0025 












.004 






1000 


.008 












.009 









It is seen that the coefficient of friction at starting is 
much greater than at stopping, and that these are both 
much greater than the value at a speed of 150 ft. per 
minute. 

For an intermittent feed such as is given by one oil hole, 
without a cup, oiled occasionally. Professor Thurston 
found for steel shaft in bronze bearings, with a speed of 
rubbing of 720 ft. per minute, the following coefficients 
of friction: 



FRICTION 



269 





Pressuke, lb. i'er sq. in. 


Oil 


8 


16 


32 


48 


Sperm and lard 
Olive and cotton seed 
Mineral oils 


.150-.25 

.1G0-.283 

.154-.2()1 


.l;38-.192 
.107-.245 
.145-.233 


.086-.141 
.101-.108 
.086-. 178 


.077-.144 
.079-.131 
.094-.222 



The results show that the coefficient decreases with the 
pressure within the range reported, but that the results 
are considerably higher than those for well-lubricated 
bearings. He also found in connection with the same 
tests that with continuous lubrication sperm oil gave the 
following coefficients; 



Pressure, 
lb. per sq. in. 

50 

200 

300 



Coefficient 
OF Frictioh. 

.0034 
.0051 
.0057 



The results of tests of the friction of well-lubricated 
bearings are summarized by Goodman (^Engineering News^ 
April 7 and 14, 1888) as follows: 

(a) The coefficient of friction of ivell-luhricated surfaces is 
from \ to Jq that of dry or poorly lubricated surfaces. 

(6) The coefficient of friction for moderate pressures and 
speeds varies approximately inversely as the normal pressure; 
the frictional resistance varies as the area in contact^ the nor- 
mal pressure remaining the same, 

(c) For low speeds the coefficient of friction is abnormally 
high^ but as the speed of rubbing increases from about AO to 
100 ft, per miiiute, the coefficient of friction diminishes^ and 
again rises when that speed is exceeded^ varyiiig ajyproxi- 
ynately as the square root of the speed. 



270 



APPLIED MECHANICS FOR ENGINEERS 



(c?) The coefficient of friction varies approximately in- 
versely as the temperature^ within certain Ihnits; namely^ 
just hefore abrasion takes place, 

149. Method of Testing Lubricants. — To make the matter 
of the tests of the friction of lubricants clear, it will be 
convenient to make use of the description of a testing 



cTBlJl 




Fig. 171 

machine used by Dean W. F. M. Goss at Purdue 
University on graphite, and a mixture of graphite and 
sperm oil. In making the tests the apparatus shown in 
Figs. 171 and 172 was used. (See "A Study in Graph- 
ite," Joseph Dixon Crucible Co.) 

This apparatus represents, in principle, the machines 
generally used for testing lubricants. It is therefore 
shown in some detail. The weight Gr is hung from the 
shaft upon which it is suspended by the form of box to be 
tested. The desired speed of rubbing is obtained by 
means of the cone of pulleys, and the pressure on the bear- 



FRICTION 



271 



R-\-G 



ing is adjusted by the spring. The temperature of the 
bearing is read from the thermometer inserted in the bear- 
ing. When rotation takes place, the 
weight Gr is rotated a certain distance 
dependent upon the friction. This dis- 
tance is measured on the scale A, The 
forces acting upon the pendulum Gr are 
shown in Fig. 172, where R represents 
the resistance of the spring, ^the force 
of friction, I the distance of the center 
of gravity of Gr from the axis of rota- 
tion, (/) the angle through which Gr is 
deflected, and r the radius of the 
shaft. Taking moments about the 
center of the shaft, we have, when Gi 
is held in the position shown, due to 

friction, 

F-^r = Gri sin (/>, where F^ is the total 

friction on the bearing 2F = F^; 




Fig. 172 



but 



so that 



^_ Grl sin (f> 



It is customary to take G small compared with i?, so 
that the pressure on both sides of the bearing may be 
considered equal to 72, the resistance of the spring. The 
spring is easily calibrated so that R may be made any- 
thing desired by compressing the spring through the 
appropriate distance as indicated on the scale F(Fig. 171). 
The quantities (r, Z, r, and It are known, and (j> can be 
read so that/ can be calculated. 



272 



APPLIED MECHANICS FOR ENGINEERS 



If G is not small compared to i2, then 



/= 

so that 



F, 



'XF, 



F, 



average pressure .XR + G-) + E 

Crl sin ^ 



B + 



a 



/= 



r[R^^]R 



The results of tests made upon a mixture of graphite 
and oil as a lubricant are given in the pamphlet. The 
tests were run under 200 lb. per square inch pressure, 
at a speed of rubbing of 145 ft. per minute. Oil was 
dropped into the bearing at the rate of about 12 drops per 
minute, showing a coefficient of friction of \, 

Problem 234. If the weight of the pendulum is 360 lb., the 
diameter of the shaft 4J in., distance of the center of gravity of G 
from the center of shaft 2 ft., the angle <^ 5 degrees, and the average 
resistance of the spring 1000 lb., find the coefficient of friction. The 
weight G should not be neglected in this case. 

150. Rolling Friction. — The resistance offered to the 
rolling of one body over another is known as rolling fric- 
tion. It is entirely different from sliding friction, and its 

laws are not so well 
understood. When a 
wheel or cylinder (Fig. 
173) rolls over a track 
the track is depressed 
\ and the wheel dis- 



v;?//;;;??????v??^??????/ 7777r7 




Fig. 173 



torted. The force P 
necessary to overcome this depression and distortion is 
known as rolling friction. 



FRICTION 273 

The forces acting on the wheel are seen from Fig. 173 
to be: P the working force, W the weight on tlie wheel, 
and M the resistance of the track or roadway to the rolling. 
This upward pressure i2 is not quite vertical, but has its 
point of application a short distance K' from the vertical. 
Its line of action passes through the center of the wheel. 
The distance jfiT' depends chiefly upon the roadway; it is 
called the coefficient of rolling friction. It is measured in 
inches and is not a coefficient of friction in the strict 
sense that / is the coefficient of sliding friction. Taking 
moments about the point of application of M^ we have, 
approximately, 

WK^ = Pr, 

SO that K' = ^, or r= 

When the track or roadway is elastic or nearly so, we 
have a condition something like that represented in Fig. 
174. The wheel 
sinks into the ma- 
terial and pushes it 
ahead, at the same 
time it comes up 
behind the wheel. 
For a portion of 
the wheel on each 
side of the point 
the roadway is simply compressed; over the remainder of 
the surface in contact, however, slipping occurs, as 
indicated by the arrows. The resultant resistance, 
however, is in front of the vertical through the center. 




274 



APPLIED MECHANICS FOR ENGINEERS 



and we have, as in the case of imperfectly elastic 
roadways, 



P = 



K^W 



It has been found hj Reynolds (see Phil. Trans. Royal 
Soc, Vol. 166, Part 1) that when a cast-iron roller rolls 
on a rubber track, the slippage, due to the elasticity of 
the track, may amount to as much as .84 in. in 34 
in. An elastic roller rolling on a hard track will roll 
less than the geometrical distance traveled by a point on 
the circumference. When the roller and tracks are of the 
same material, the roller rolls through less than its geo- 
metrical distance. 

151. Friction Wheels. — The friction of bearings is often 
made much less by the use of friction wheels. The ar- 
rangement is usually something like that shown in Fig. 




Fig. 175 



FRICTION 275 

175. The two friction wheels A^ carry the shaft of the 
mechanism A, The friction of the shaft A is thus changed 
from sliding to rolling friction. Let P be the normal 
pressure on the shaft, and let the two equal forces P' act- 
ing through the centers of the friction wheels be the com- 
ponents of P acting on the friction wheels. The forces 
acting on each friction wheel are, then, the pressure P\ 
the friction jP, and the friction of the bearing F'. Since 
P, P\ and P' form a balanced system of forces, when the 
forces acting on the shaft are considered, 

2cos/3' 

where 2/8 is the angle between the forces P' . The value 
of the friction F' of the friction wheel bearings is 

cos p 

where/ is the coefficient of sliding friction, and the moment 
of this friction is 

cosp 

Taking moments about the center of a friction wheel, we 

have 

Fr^ = PV3, 
so that 

cosp 



^=(?^-^- 



It is seen that if the ratio -^ is constant, the friction 

may be made less by taking yS small, so that cos/3 is large. 
If r^ is large as compared with rg, the friction is reduced. 



276 APPLIED MECHANICS FOB ENGINEERS 

Tlie work lost due to friction per revolution of A is 
2 Trr^ F, or 

^ \7\y J COS P 



The friction of A when resting in an ordinary bearing 
would be /P. In order that the friction of the friction 
wheels may be less than that of a plain bearing, we must 
have 



r, 



^-— <1, or -^<cos^. 



ro cos yS r, 



2 



The work lost, per revolution of J., in a plain bearing, 
would be 2 irr^fP, It is seen that the criterion that the 
work lost in tlie friction bearing be less than that lost in 
the plain bearing is the same as that given above, viz.. 



cos)8> 



!3 
^2 



If the angle /3 is zero, that is, if there is only one fric- 
tion wheel, so that the center of A is vertically over J.', 
the friction is 

^2 

This is always less than the friction of a plain bearing, 
since -^ is always less than unity. 

^2 

Problem 235. li P— ^ tons and the radius of the shaft is 2 in. 
and the coefficient of friction is .07, what work is lost per revolu- 
tion ? If the shaft makes 3 revolutions per second, what horse power 
per revolution is lost in friction? Given also /3 = 45°, rg = | in., and 
Tg = 4 in. 



FRICTION 277 

Problem 236. In the case of the shaft nieutioiied in the preced- 
ing problem, how much more horse power per revohition would it 
take if the bearing was plain ? What value of jB would give the 
same loss due to friction in both the plain bearing and the one pro- 
vided with friction wheels ? 

152. Resistance of Ordinary Roads. — Resistance to trac- 
tion consists of axle friction, rolling friction, and grade 
resistance. Axle friction varies from .012 to .02 of the 
load, for good lubrication, according to Baker. The 
tractive power necessary to overcome axle friction for 
ordinary American carriages has been found to be from 
3 lb. to 3| lb. per ton, and for wagons with medium-sized 
wheels and axles from 3| lb. to 4 J lb. per ton. 

The total tractive force per ton of load, for wheels 50 in., 
30 in., and 26 in., in diameter, respectively, is, according 
to Baker (^Engineering Netvs^ March 6, 1902) : 





Tractive Force 
IN Pounds 


On macadam roads 

On timothy and blue grass sod, dry, grass cut 
On timothy and blue grass sod, wet and springy . 
On plowed ground, not harrowed, dry and cloddy 


57 
1:32 
173 
252 


61 
145 
203 
303 


70 
179 
288 
374 



Rolling resistance is influenced by the width of the tire. 
According to Baker, poor macadam, poor gravel, compres- 
sible earth roads, and, on agricultural lands, narrow tires, 
usually give less traction. On earth roads composed of 
dry loam with 2 to 3 in. of loose dust, traction with 
l|-in. tires was 90 lb. per ton, and with 6-in. tires 106 
lb. per ton. On tlie same road when it was liard and 
dry, with no dust, that is, wlien it was compressible, the 



278 



APPLIED MECHANICS FOR ENGINEERS 



traction was found to be 149 lb. per ton with l|^-in. tires 
and 109 lb. per ton with 6-in. tires. On broken stone 
roads, hard and smooth, with no dust or loose stones, the 
traction per ton was 121 lb. with l|-in. tires, and 98 lb. 
with 6-in. tires. Moisture on the surface or mud in- 
creases the traction. 

Morin found that with 44-in. front and 54-in. rear 
wheels on hard dry roads the traction per ton was 114 lb. 
with either l|-in. or 3-in. tires. On wood-block pave- 
ments the traction per ton was 28 lb. with 1^-in. tires, 
and 88 lb. with 6-in. tires. 

On asphalt, bricks, granite, macadam, and steel-road sur- 
faces, investigated by Baker, the traction per ton varied 
from 17 lb. to 70 lb., the average being 38 lb. 

Morin gives the coefficient of rolling friction for wagons 
on soft soil as .065 in., and on hard roads .02 in. Accord- 
ing to Kent ("Pocket-Book"), tests made upon a loaded 
omnibus gave the following results : 



Pavement 


Speed, Miles 
PER Hour 


Coefficient, 
Inches 


Eesistance, per 
Ton, in Lb. 


Granite 

Asphalt 

Wood 

Macadam, graveled . . 
Macadam, granite, new . 


2.87 
3.56 
3.34 
3.45 
3.51 


.007 

.0121 

.0185 

.0199 

.0451 


17.41 
27.14 

41.60 

44.48 

101.09 



Problem 237. Compare tlie resistance offered to a load of two 
tons pulled over asphalt, macadam, good earth roads, or wood-block 
pavement. Width of tires, 6 in. 

Problem 238. Compare the resistances in the above problem with 
that of steel rails to the same load. 



FRICTION 



279 



153. Roller Bearings. — In the roller bearings the sliaft 
rolls on hardened steel rollers as shown in cross section in 
Fig. 176. The roll- 
ers are kept in place 
in some way similar 
to that shown in the 
journal of Fig. 177. 
Such bearings are 
used where heavy 
loads are to be car- 
ried. Tests of roller 
bearings have been made by Dean C. H. Benjamin 
(^Machinery^ October, 1905), who determined the follow- 
ing values for the coefficient of friction. Speed 480 
revolutions per minute. 




Fig. 176 





Diameter of 


Roller Bearing 


Plain Cast-Iron Bearing 


Journal, in Inches 


Max. 


Min. 


Average 


Max. 


Min. 


Avera«^e 


2tV 
211 


.036 
.052 

.041 
.053 


.019 

.034 
.025 
.049 


.026 
.040 
.030 
.051 


.160 
.129 

.143 
.138 


.099 

.071 
.076 
.091 


.117 
.094 
.104 
.104 





It was found that the coefficient of friction of roller 
bearings is from ^ to |- that of plain bearings at moderate 
speeds and loads. As the load on the bearing increased, 
the coefficient of friction decreased. Tightening the bear- 
ing was found to increase the friction considerably. 

Tests of the friction of steel rollers 1, 2, 3, and 4 in. 
in diameter are reported in the Transactions Am. Soc. 
C. E., August, 1894. The rollers were tested between 



280 



APPLIED MECHANICS FOR ENGINEEBS 



plates IJ in. thick and 5 in. wide, arranged as shown in 
Fig. 178. Tests were made with the plates and rollers of 




Fig. 177 




^P' 



cast iron, wrought iron, and steel. The friction P' for 

..1^7. ^ A ^ I. -0063 

unit load P was round to be 



Vr 



for cast-iron rollers 
.0073 



and plates, ^^^ for wrought iron, and '——- for steel, 
Vr vr 

when r represents the radius of 

the roller in inches. The rollers 

were turned and the plates 

planed, but neither were polished. 




154. Ball Bearings. — For high 
speeds and light or moderate 
loads the friction is much re- 
duced by the use of hardened 
Fig. 179 steel balls instead of the steel 

rollers. These bearings are now used on all classes of 



FRICTION 



281 



machinery, giving a much greater efficiency except for 
heavy loads. The principal objection to the ball bearing 
seems to be clue to the fact that there is so little area of 
contact between the balls and bearing plates. This gives 
rise to very high stresses over these areas, and consequently 
a considerable deformation of the balls. When the ball 
has been changed from its spherical form it is no longer 
free to roll, and the friction increases rapidly. Some 
authorities consider a load of from 60 to 150 lbs. sufficient 
for balls varying in size from | to -| inch in diameter. 
Figure 179 illustrates a type of bearing used for shafts, 
and Fig. 180 a type used for 
thrust blocks. 

The conclusions reached by 
Goodman from a series of tests 
on bicycle bearings (Proc. Inst. 
C. E., Vol. 89) are as follows: 

(1) The coefficient of friction 
of ball hearings is constant for 
varying loads, hence the frictional 
resistance varies directly as the 
load. 

(2) The friction is unaffected hy a change of temperature. 
The bearings were oiled before starting the tests. The 

coefficient of friction for ball bearings Avas found to be 
rather higher than for plain bearings with bath lubrica- 
tion, but lower than for ordinary lubrication. Ball bear- 
ings will also run easily with a less supply of oil. The 
following table gives the resvilts of tests of ball bearings. 
The bearings were oiled before starting, and the tests Avere 
run at a temperature of 68° F. 




Fig. 180 



282 



APPLIED MECHANICS FOB ENGINEERS 





19 




is: 


r 


350 


Load on 
Bearing 


Retolutions 


PER MiN. 


Revolutions 


PER MiN. 


Revolutions per Min. 


IN Lb. 


Coetf. friction 


Friction, lb. 


Coeff. friction 


Friction, lb. 


Coeff. friction 


Friction, lb. 


10 


.0060 


.06 


.0105 


.10 


.0105 


.10 


20 


.0045 


.09 


.0067 


.13 


.0120 


.24 


30 


.0050 


.15 


.0050 


.15 


.0110 


.33 


40 


.0052 


.21 


.0052 


.21 


.0097 


.39 


50 


.0054 


.27 


.0054 


.27 


.0090 


.45 


60 


.0050 


.30 


.0055 


.33 


.0075 


.45 


70 


.0049 


.34 


.0054 


.38 


.0068 


.47 


80 


.0048 


.38 


.0062 


.49 


.0060 


.48 


90 


.0050 


.45 


.0068 


.61 


.0060 


.54 


100 


.0058 


.58 


.0069 


.69 


.0057 


.57 


110 


.0054 


.59 


.0065 


.71 


.0060 


.66 


120 


.0055 


.66 


.0075 


.90 


.0057 


.68 


130 


.0058 


.75 


.0078 


1.01 


.0062 


.81 


140 


.0056 


.78 


.0077 


1.08 


.0060 


.84 


150 


.0060 


.90 


.0083 


1.24 


.0062 


.93 


160 


.0075 


1.20 


.0081 


1.29 


.0058 


.93 


170 


.0079 


1.34 


.0078 


1.33 


.0055 


.93 


180 


.0079 


1.42 


.0078 


1.40 


.0053 


.95 


190 


.0087 


1.65 


.0076 


1.44 


.0054 


1.03 


200 


.0090 


1.80 


.0081 


1.62 


.0060 


1.20 



Another series of tests, run with a constant load on the 
bearing of 200 lb. and a temperature of 86^ F., shows the 
variation of the coefficient of friction with the speed. It 
is seen that as the speed increased the coefficient and the 
friction decreased. The preceding table, however, shows, 
for loads below 175 Zi., an increase in the coefficient with 
increase in speed. In particular, this table shows that for 
loads below 80 lbs. the coefficient increased with increase 
of speed ; for loads between 90 and 175 lbs. it increased 
when the speed was 150 r.p.m. and decreased when it was 
350 R.P.M. Beyond 175 lbs. the coefficient increased. 



FRICTION 



283 



Revolutions per Minitk 


Coefficient Friction 


Friction Pounds 


15 


.00735 


1.47 


93 


.004(35 


.93 


175 


.00375 


.75 


204 


.00345 


.69 


280 


.00300 


.60 



It seems from the data given that the first conclusion 
of Goodman's should be changed to read : the coefficient of 
friction of hall hearings is constant for vari/ing loads^ up to 
a certain limits heyond which it increases with increase of 
load. This limit is about 150 lb. in the tests reported. 

Tests on ball bearings designed for machinery subjected 
to heavy pressures have been made in Germany (see Zeit- 
schrift des Vereins deutsche Ingenieure^ 1901, p. 73). It 
was found that at speeds varying from 65 to 780 revolu- 
tions per minute, vt^here the bearing was under pressures 
varying from 2200 lb. to 6600 lb., the coefficient of friction 
varied little and averaged .0015. 

Tests of ball bearings made by Stribeck and reported by 
Hess (Trans. Am. Soc. M. E., Vol. 28, 1907) give rise to 
the following conclusions : (a) the load that may be put 
upon a bearing is given by the formula 



r = 



cd-n 



where P is the load in pounds on a bearing, consisting of 
one row of balls, (? is a constant dependent upon the mate- 
rial of the balls and supporting surfaces and determined 
experimentally, d the diameter of the balls, the unit being 
^ of an inch, and n the number of balls. For modern 



284 



APPLIED MECHANICS FOR ENGINEERS 



materials c varies from 5 to 7.5. (5) The coefficient of 
friction varied from .0011 to .0095. It was independ- 
ent of speed, '' within wide limits," and approximated 
.0015; this was increased to .003 when the load was 
about one tenth the maximum. 

The following values for the coefficient of friction for 
heavy loads are reported, from observation, with the state- 
ment that the real values are probably somewhat less: 



Revolutions 
per minute 


65 


100 


190 


380 


580 


780 


1150 


Coefficient of 
















friction for 
















load 840 lb. 


.0095 


.0095 


.0093 


.0088 


.0085 




.0074 


Coefficient of 
















friction for 


♦ 














load 2400 lb. 


.0065 


.0062 


.0058 


.0053 


.0050 


.0049 


.0047 


Coefficient of 
















friction for 
















load 4000 to 
















9250 lb. 


.0055 


.0054 


.0050 


.0050 


.0041 


.0041 


.0040 



It should be remembered that the friction of a ball 
bearing is due to both sliding and rolling friction, the 
sliding friction being due to the elasticity of the balls 
and the bearing (see Art. 150). Rolling friction is most 
nearly approached when the balls are hard and not easily 
changed from their spherical shape. All materials, how- 
ever, are deformed under pressure so that perfect rolling 
friction is impossible. On account of the sliding friction 
present in roller and ball bearings, it is necessary to use a 
lubricant to prevent wear. 



FRICTION 



285 



Problem 239. ITow many J-iii. balls will be necessary in a ball 
bearing designed to carry 4000 lb., if c = 7.5? If /= .0015, what 
work is lost per revolution, the distance from the axis of rotation to 
the center of balls being one inch? 

155. Friction Gears. — In the friction gears the driver 
is usually the smaller wheel, and when there is any differ- 
ence in the materials of which tlie wheels are made, the 
driver is made of the softer material. This latter arrange- 
ment is resorted to, to prevent flat places being worn on 
either wheel in case of slipping. These gears have been 
used for transmitting light loads at high speeds, where 
toothed gears would be very noisy, or in cases where it is 
necessary to change the speed or direction of the motion 
quickly. 



IRON FOLLOWER 




Fig. 181 



The use of paper drivers has made possible the trans- 
mission of much heavier loads by means of such gears. 



286 APPLIED MECHANICS FOB ENGINEERS 

A series of tests, made by W. F. M. Goss, and reported 
ill Trans. Am. Soc. M. E., Vol. 18, on the friction be- 
tween paper drivers and cast-iron followers, is of inter- 
est in this connection. The apparatus used is shown in 
Fig. 181. The pressure between the wheels was obtained 
by a mechanism that forced the two wheels together w^itli 
a pressure P. A brake w^heel shown in the figure ab- 
sorbed the power transmitted. 

The coefficient of friction was regarded as the ratio of 
F to P, as in sliding friction. While this is customary, 
it is not entirely true, since we have the rolling of one 
body over the other. We shall, however, assume that we 

may call the coefficient of friction /=—• It was found 

that the coefficient of friction varied with the slippage, 
but was fairly constant for all pressures up to some point 
between 150 to 200 lb. per inch of width of wheel face. 
'' Variations in the peripheral speed between 400 and 2800 
ft, per minute do not affect the coefficient of friction.''^ 

If the allowable coefficient of friction be taken as .20, 
the horse power transmitted per inch of width of face of 
the wheel is 

H.P. = 150 X .2x^\7rd xwx ^^M0288dwl^, 

33,000 

where d is the diameter of the friction wheel in inches, 
w the width of its face in inches, and iV the revolutions 
per minute. Using this formula, the following table is 
given in the article in question : 



FRICTION 



287 



Horse Power which may be Transmitted by Means of Paper 

Friction Wheel of One Inch Face, when run 

UNDER A Pressure of 150 lb. 



DiAMETEK 




Kf.volutions per M 


[NUTE 




r\v Pitt t w 












IN Inches 


25 


50 


75 


100 


150 


200 


600 


1000 


8 


.0476 


.0952 


.1428 


.1904 


.2856 


.3808 


1.1424 


1.904 


10 


.0595 


.1190 


.1785 


.2380 


.3570 


.4760 


1.4280 


2.380 


14 


.0833 


.1666 


.2499 


.3332 


.4998 


.6664 


1.9992 


3.332 


16 


.0952 


.1904 


.2856 


.3808 


.5712 


.7616 


2.2848 


3.808 


18 


.1071 


.2142 


.3213 


.4284 


.6426 


.8568 


2.5704 


4.288 


24 


.1428 


.2856 


.4284 


.5712 


.8568 


1.1424 


3.4272 


5.712 


30 


.1785 


.3570 


.5355 


.7140 


1.0710 


1.4280 


4.2840 


7.140 


36 


.2142 


.4284 


.6426 


.8568 


1.2852 


1.7136 


5.1408 


8.560 


42 


.2499 


.4998 


.7497 


.9996 


1.4994 


1.9992 


5.9976 


9.996 


48 


.2856 


.5712 


.8568 


1.1424 


1.7136 


2.2848 


6.8544 


11.420 



The value of the coefficient of friction for friction gears, 
(Kent, ''Pocket Book") may betaken from .15 to .20 for 
metal on metal; .25 to .30 for wood on metal; .20 for 
wood on compressed 
paper. 

Problem 240. If the 

friction wheels are grooved 
as shown in Fig. 182, both 
of cast iron, and the small 
driver fits into the groove 
of the larger follower, we 
may take/= .18. Then 

F = 2fN = 2fP cos a 

= .36 P cos a. Fig. 182 

Problem 241. The speed of the rim of two grooved friction 
wheels is 400 ft. per minute. If a = 45°, /= .18, what must be the 
pressure P to transmit 100 horse power ? 




288 



APPLIED MECHANICS FOR ENGINEERS 



Problem 242. What horse power may be transmitted by the 
gearing in the preceding problem, ii P = 6000 lb. and the peripheral 
velocity is 12 ft. per second? 

156. Friction of Belts. — When a belt or cord passes over 
a pulley and is acted upon by tensions T^ and T^^ the ten- 
sions are unequal, due to the friction of the pulley on the 
belt. We shall determine the relation between T^ and T^. 
Let the pulley be represented in Fig. 183. The belt covers 




Fig. 183 



an arc of the pulley whose angle is a. Consider the forces 
acting upon the belt and suppose 2\ and T^ to be the ten- 
sions in the belt on the tight and slack sides, respectively, 
and T the tension in the belt at any point of the arc of 
contact. Let F be the total friction between the pulley 
and belt and dF the friction on an elementary arc ds. If 
dp is the noi'mal pressure on an elementary arc, then 
dF=fdP and T^- T.^ = F, where / is the coefficient of 
friction. 

Represent as in Fig. 183 an elementary arc of the belt 



FRICTION 



289 



of length ds. The forces acting on this elementary part 
are, the tensions T + dT and T^ the friction dF^ and the 
normal pressure dP. Taking moments about the center 
of the pulley, we have 

(^T+dTy = dFr+Tr, 
or dT=dF. 

Of the forces acting upon this elementary portion of the 
belt, dT and dF are in 
equilibrium, so that T^ dP^ 
and T must also be in equi- 
librium. Since this is true, 
these latter forces must 
form a closed triangle when drawn to scale (Art. 13). 
We have, then, from Fig. 184, approximately, 

dP= TJ/3, 

dT=dF=fTdl3, 




Fig. 184 



SO that 



or 



This gives 



or 



*^T2 I *^0 

T 



or, writing it in the exponential form, 

This is the relation desired. The quantity e = 2.72+ is 
the base of the system of natural logarithms. The 



log 



10 



.4343. 

F=T,^T,^=T,(l 
u 



-fo. 



) 



T./y'^ 



1) 



290 



APPLIED MECHANICS FOB ENGINEEES 



When the band is used to resist the motion of a pulley- 
as in some types of brakes (see Fig. 185), it is known as a 
friction strap, see Art. 165. 




Fig. 185 



Problem 243. A rope is wrapped four times around a post and a 
man exerts a pull of 50 lb. on one end. If the coefficient of friction is 
.3, what force can be exerted upon a boat attached to the other end 
of the rope ? 

Problem 244. A pulley 4 ft. in diameter, making 200 revolutions 
per minute, drives a belt that absorbs 20 horse power. What must be 
the width of the belt in order that the tension may not exceed 70 lb. 
per inch of width ? 

Problem 245. What should be the width of a belt J of an inch 
thick to transmit 10 horse power ? The belt covers .3 the smaller 
pulley and has a velocity of 500 ft. per minute. The coefficient of fric- 
tion is .27 and the strength of the material 300 lb. per square inch. 

Note. The power that can be transmitted by a belt depends 
upon the friction between the belt and pulley. So that 



H.P. 



Fv _ (T,-T,)v 



33,000 



33,000 



FRICTION 



291 



'V 157. Transmission Dynamometer. — It has been shown, 
Art. 156, that the tension 



in 




I 



of a belt on the tight side is 
greater than the tension on the 
slack side. The transmission dy- 
namometer (the Fronde dynamom- 
eter), illustrated in Fig. 186, is 
designed to measure the difference 
in these tensions. Let the pulley 
D be the driver and the pulley E T, 
the follower, so that 7\ represents 
the tight side of the belt and T^ 
the slack side. The pulleys B^ B 
run loose on the T-shaped frame 
CBB, This frame is pivoted at A. 
If we neglect the friction due to 
the loose pulleys, we have the fol- 
lowing forces acting on the T- 
frame, two forces T^ at the center 
of the right-hand pulley B^ two 
forces T^ at the center of the left- 
hand pulley jB, a measurable re- 
action P at (7, and the reaction of the pin at A, Taking 
moments about the pin, we have 

P((7^) = 2 T^iBA) - 2 T^(^BA} 

= 2BAiT,-T,\ 




Fig. 18(3 



so that 



^ - 2 BJL 



The distances CA and BA are known, and P may be 
measured ; the difference, then, T^ — T^, may always be 



292 APPLIED MECHANICS FOR ENGINEERS 

obtained. The value T^ — T^ is then known and the horse 
power determined by the rehition (see Problem 253) 

' ' ' 33,000 

where n is the number of revolutions, r is the radius of the 
machine pulley in feet. 

158. Creeping or Slip of Belts. — A belt that transmits 
power between two pulleys is tighter on the driving side 
than it is on the foUoiving side. On account of this differ- 
ence in tension and the elasticity of the material, the tight 
side is stretched more than the slack side. To compen- 
sate for this greater stretch on one side than on the other, 
the belt creeps or slips over the pulleys. This slip has 
been found for ordinary conditions to vary from 3 to 12 
ft. per minute. The coefficient of friction when the slip 
is considered is about .27 (Lanza). It has also been 
found that the loss in horse power in well-designed belt 
drives, due to slip, does not exceed 3 or 4 per cent of the 
gross power transmitted, and that ropes are practically 
as efficient as belts in this respect. For an account of the 
experimental investigations on this subject the student is 
referred to Institution of Mechanical Engineers, 1895, 
Vols. 3-4, p. 599, and Transactions Am. Soc. M. E., Vol. 
26, 1905, p. 584. 

159. CoeiScient of Friction of Belting. — The value of 
the coefficient of friction of belting depends, not only on 
the slip but also upon the condition and material of tlie 
rubbing surfaces. Morin found for leather belts on iron 
pulleys the coefficient of friction /= .50 when dry, .30 



FRICTION 



293 



when wet, .23 when greasy, and .15 when oily (Kent, 
''Hand Book"). Most investigators, however, including 
Morin, took no account of slip, so that the best value of/, 
everything considered, is that given in the preceding 
article (.27). 

160. Centrifugal Tension of Belts. — When a belt runs at 
a high rate of speed over a pulley there is considerable 
tension introduced in the belt due to the centrifugal force. 
We have seen (Art. 86) that the centrifugal force equals 

■ , where 31 is the mass and v tlie tangential velocity. 

Let the centrifugal force be represented by P^ and the 

tension in the belt due to this force by T^.. We know 

3Iv^ 

that Pc = • Now if we consider a section of belt one 

r 

foot long and of one square inch cross section, we may 

consider the tensions Tc, at either end of this len^^th, in 





Fig. 187 



equilibrium with P^ (see Fig. 187 (^) ). From Fig. 187 
(5) we have approximately P^ = T^.6, but from Fig. 187 

(a), 6=-, so that P^ = —^ Since Pc = — -, 
r r. r 



294 



APPLIED MECHANICS FOR ENGINEERS 



T = 31 v'- = 



9 



where W is the weight of a portion one foot long and one 
square inch in cross section. If 7 for leather is 56 lb., 

Tr= .388 lb. and T,= -~v'- = .012 v^. 

Hence, in designing belts, the total tension must be 
T^+T,^ ^i + .012z;2= T^e^-+,012v\ 

Problem 246. A belt runs at a velocity of 4000 ft. per minute. 
What tension is introduced by the centrifugal force of the belt in 
passing over the wheel ? 

Problem 247. What additional width of belt must be provided 
for in Problems 244 and 245 if the centrifugal force of the belt is 
considered ? 

161. Stiffness of Belts and Eopes. — Belts and ropes used 

in the transmission of power 
are not perfectly flexible, so 
that some force is necessary to 
bend them around the pulleys. 
We desire to know the magni- 
tude of this force. Let T 
(Fig. 188) be the tension in 
the on-side of the belt and 
T + T^ the tension on the 
off-side. Neglecting the effect 
of the friction of the pulley, T^ 
represents the force necessary 
to overcome the stiffness of 
tlie rope. In the analysis here 
Fig. 188 * ' ^' given, it is assumed that while 




r+r. 



FRICTION 295 

it takes a certain force T^ to cause the rope to wind around 
the pulley, it does not require any force to straighten it. 
This assumption is nearly true for steel wire rope, but not 
nearly so true for hemp rope. All rope requires some 
force to straighten it when coming off the pulley. 

Taking moments about the center of the pulley and 
neglecting the friction of the bearing, we have 

or T^ = y(^^ ~ ^i\ 

where d^= r + -, and d^ = r + ai+- + a^- 

The distance a^ is due to the stiffness of the rope, and 
the distance a^ the distance of the point of application of 
T from the center of the rope. That T does not act at 
the center of the rope, but at a distance a^ toward the 
outside, is due to the fact that the outside of the roj)e is 
under greater tension than the inside. Now the distance 
a^ for inelastic ropes decreases as T increases, and so we 

may write a^ = ^, where c^ is a constant, determined 

experimentally. For wire rope, a^ increases with in- 
creased radius of the pulley, and decreases with increased 
tension, so that we may write 



.(-f) 



«i = Y 



making these substitutions in the above equation, we have 

e^ + d^T 
1\ = — -1 — tor hemp rope, 



296 APPLIED MECHANICS FOR ENGINEERS 

and a^T , 

2\ = c^-i ^— y tor wire rope. 

For tarred hemp ropes, c-^ has been found (see Du Bois, 
''Mechanics of Engineering") to be 100, and a^^ .222, so that 

^ 100 + .222y . 

^^ = \ — pounds. 

, Cv 

For new hemp ropes^ 

T^ = —ZLl — _ pounds. 

For wire ropes^ 

m i Ao .0937 7 , 

y=1.08H -J- pounds. 

^ + 2 

In each case T is expressed in pounds and r and d in 
inches. 

Problem 248. — A new hemp rope, one inch in diameter, passes 
over a pulley 13 in. in diameter, under a tension of 500 lb. What 
is the force necessary to overcome the stiffness of the rope? What 
per cent is this of the total tension in the rope ? 

Problem 249. — A wire rope, one inch in diameter, passes over a 
pulley 2 ft. in diameter, under a tension of 1000 lb. What force 
is necessary to overcome the stiffness of the rope? Compare this force 
with that necessary to overcome the stiffness of the same rope under 
the same tension, when it passes over a pulley 12 in. in diameter. 
What per cent of the total tension is it in each case ? 

Problem 250. — A new hemp rope, 2^ in. in diameter, passes 
over a grooved pulley 31 in. in diameter, under a tension of 1000 
lb. What force is necessary to overcome the stiffness of the rope? 
Allow an increase of 6 per cent for the grooved pulley. 



FRICTION 297 

It will be seen from the formula and the problems tliat 
the force necessary to overcome the stiffness of ropes is 
greater for small pulleys than for large ones. 

The following empirical formula will be found useful 
(see Memoirs et Compte rendu de la Societe des Ingenieurs 
Civile^ December. 1893, p. 558, or Proc. Inst. C. E., Vol. 
116, p. 455): 

for ropes, where c?and r are expressed in millimeters and 
2\ and T in kilograms. A formula for belting is also 
given, 



t, = ^^^m-^ut']. 



where w is the width of the belt and t its thickness. T^ 
and T are expressed in kilograms and w^ t^ and r in milli- 
meters. 

The student should solve Problems 248 and 249, using 
the empirical formula given above. 

Problem 251. A belt 12 in. wide and -| in. thick passes over 
a pulley 18 in. in diameter under a tension of 1000 lb. What 
force is necessary to overcome the stiffness of the belt? 

In using the empirical fornmla just given it will be necessary to 
change pounds to kilograms and distances to millimeters. 

162. Friction of a Worn Bearing. — Tlie friction of a bear- 
ing that fits perfectly is the friction of one surface sliding 
over another and is given by the equation 



298 



APPLIED MECHANICS FOR ENGINEERS 



where F is the force of friction, / the coefficient of friction, 
and iV'is the total normal pressure on the bearing. 

When, however, the bearing is worn, as is shown much 
exaggerated in Fig. 189, the friction may be somewhat 




Fig. 189 

different. When motion begins, the shaft will roll up on 
the bearing until it reaches a point A where slipping be- 
gins. If motion continues, slipping will continue along a 
line of contact through A. Let P be a force that causes 
the rotation, R a force tending to resist the rotation, and 
M^ the reaction of the bearing on the shaft. There are 
only three forces acting on the shaft, so that P, i2, and R^ 
must meet in the point B. The direction of R^ is accord- 
ingly determined. The normal pressure is iV^= i^^ cos ^, 
and the force of friction is 

It is seen that 6 is the angle of friction. The moment of 
the friction with respect to the center of the axle is 



Fr 



m^r sin 6. 



FRICTION 299 

If the axle is well lubricated so that 6 is small and sin 6 
may be replaced by tan 6 = f , the friction is 

and the moment 

Fr = fM^r. 

The circle tangent to AB radius r sin is called the 
friction circle. Since r and 6 are known generally, this 
circle may be made use of in locating the point A, In 
other words, the shaft will continue to rotate in the bear- 
ing so long as the reaction B^ falls within the friction 
circle, and slipping will begin as soon as the direction of 
B^ becomes tangent to the friction circle. 

Problem 252. If the radius of the shaft is 1 in., = 4°, 
P = 500 lb., ai = 3 ft., a-z = 2 ft., angle between ai and ao is 100° and 
P and R are at right angles to ai and a2, what resistance R may be 
overcome by P when slipping occurs ? 

Problem 253. The radius of a shaft is 1 in., R = 20 lb., 
P = 20 lb., ai = 3 ft., and a2 = 2 ft. What force of friction will be 
acting at the point A, when the angles between P and ai and R and 
ao are right angles ? What must be the value of the coefficient of 
friction ? 

163. Friction of Pivots. — The friction of pivots presents 
a case of sliding friction, so that the force of friction F 
equals the coefficient of friction times the normal pressure. 
That is, 

(a) Flat- End Pivot. — The friction on a flat-end pivot 
is greatest on the outside and varies linearly to zero at the 
center as shown in Fig. 190. The resultant force of friction 



300 



APPLIED MECHANICS FOR ENGINEERS 



R has its point of application |r from the center. We 

may write 

F=R=fP, 

and the moment of the friction with respect to the center 

is 

Moment = | rfP. 

p 




/^ — > 



^ ^ 

Fig. 190 Fig. 191 

The work lost per revolution is 

(6) Collar Bearing or Hollow Pivot. Let the outside 
radius be r^ and the inside radius r^ (Fig. 191); then, 



F=B=.fP, 



FRICTION 



301 



and the moment of friction is 



Moment = 



2rg- 



fP 



:i rl - rl 

(For the position of the center of gravity, see Art. 24.) 
The work lost, due to friction, per revolution is 

3L.2 



W 



•» 1 



r/P. 



( "^'^^^i^ 



If r^ = 0, this reduces to the work lost per revolution 
for the solid flat-end pivot. 

((?) Conical Pivot. 
The conical pivots, 
such as are illustrated 
in Fig. 192, do not 
usually fit into the 
step the entire depth 
of the cone. Let the 
radius of the cone at 
the top of the step be 
/, a half the angle of 
the cone, and P^ the 
resultant normal re- 
action of the bearing 
on the pivot. Then 




Fig. 192 






z Sin a 



and the total friction F= li — 



fP 



i sin a 
The moment of friction in this case is 

fp 



Moment = 






2 sin a 



302 



APPLIED MECHANICS FOR ENGINEERS 



since the resultant friction may be regarded as applied at 
I r'. The work lost due to friction per revolution may 
be written - ^-o 

3 sin a 

If cc = — , this value for work lost reduces to the work 

lost per revolution, in the case of the flat-end solid pivot. 
It is easily seen, since sin a is less than unity, that if r' is 
nearly equal to r, the friction of the conical bearing is 
greater than the friction of the flat-end bearing. This 
might have been expected from the wedgelike action of 
the pivot on the step. It is also easily seen that r' may 
be taken small enough so that the friction will be less 
than the friction of the flat pivot. The work lost due to 
friction in the case of the conical pivot will be equal to, 
greater, or less than, the work lost, due to friction in the 

case of the flat-end pivot, 
p according as 



r'-^r sin a. 

(c?) Spherical Pivot, 
Suppose the end of the 
pivot is a spherical sur- 
face, as shown in Fig. 
193. Let r be the radius 
of the shaft and r^ the 
radius of the spherical 
surface; then the load 
per unit of area of 

P 




horizontal surface is 



Fig. 193 



irr' 



FRICTION 303 

The horizontal projection of any elementary circle of the 
bearing, of radius x^ is 2 irxdx. The load on this area is 

\7rrv ^" 

and the corresponding normal pressure is 

dF, = ?Zf^ sec /3. 



X. ^ r, OD Vr^-a^ ^\.^;jT> ^Pxdxf r, \ 
But cos/3= — = — ^ , so that dF.= — ( — — ^ — ]• 

^1 ^1 r^ Wrf-x^J 

The corresponding friction is/dP^, 

Integrating between the values x = and x = r^ the 
value for tlie total friction is given by 

F=E= n-Pnf( ^dx \ 
Jo r2 V Vrf - xy 



Since r = r^ sin a and Vr^ — r^= r^ cos a, the expression 
for the friction may be written 



1 + cos CI 

IT 



If « = ^9 that is, if the bearing is hemispherical, 

Li 

F=^fPy and if «=0, that is, if the bearing is flat, 
F=fP. 

The moment of the friction is given by adding all the 
tQvm^fdP^x by means of integration; this gives 

Moment =:M^f'l sin"!- - ?! V/f=^\ 
r^ \'l r^ 2 / 

or in terms of a. 

Moment = fBr ( — V " ^^^ ^) ' 



304 APPLIED MECHANICS FOR ENGINEERS 

The work lost due to friction, for each revolution, is 
found by adding the work lost by friction on each ele- 
mentary area of the bearing ; that is, by finding the sum 
of such terms as 2 irxfdP-^ by means of integration ; this 
gives 

TF= 2 ir/Fr i-^ - cot (i\ . 

If the bearing is hemispherical, a = - , and the moment 
becomes 

Moment ->^^^ 
2 

and the work lost per revolution 

The friction of flat pivots is often made much less by 
forcing oil into the bearing, so that the shaft runs on a 
film of oil. In the case of the turbine shafts of the Niag- 
ara Falls Power Company (see Art. 135) the downward 
pressure is counteracted by an upward water pressure. 
In some cases the end of a flat pivot has been floated on 
a mercury bath. This reduces the friction to a minimum 
(see Engineering^ July 4, 1893). 

The Schiele pivot is a pivot designed to wear uniformly 
all over its surface. The surface is a tractrix of revolu- 
tion; that is, the surface formed by revolving a tractrix 
about its asymptote. Its value as a thrust bearing is 
not as great as was first anticipated (see American Ma- 
chinist, April 19, 1891). 

The coefficient of friction for well-lubricated bearings 
of flat-end pivots has been found to vary from .0041 to 
.0221 (see Proc. Inst. M. E., 1891). For poorly lubri- 



FRICTION 305 

cated bearings the coefficient may be as high as .10 or 
.25 for dry bearings. 

Problem 254. Show that the work lost per revolution for the 
hemispherical pivot is 2.35 times the work lost per revolution for the 
flat pivot. 

Problem 255. The entire weight of the shaft and rotating parts 
of the turbines of the Niagara Falls power plant is 152,000 lb., tlie 
diameter of the shaft 11 in. If the coefficient of friction is con- 
sidered as .02 and the bearing a flat-end pivot, what work would be 
lost per revolution due to friction ? 

Problem 256. A vertical shaft carrying 20 tons revolves at a 
speed of 50 revolutions per minute. The shaft is 8 in. in diameter 
and the coefficient of friction, considering medium lubrication, is 
.08. What work is lost per revolution if the pivot is flat? AVhat 
horse power? What horse power is lost if the pivot is hemispherical ? 

Problem 257. What horse power would be lost if the shaft in 
the 2^receding problem was provided with a collar bearing 18 in. out- 
side diameter instead of a flat-end block? Compare results. 

Problem 258. A vertical shaft making 200 revolutions per minute 
carries a load of 20 tons. The shaft is 6 in. in diameter and is 
provided with a flat-end bearing, well lubricated. If the coefficient 
of friction is .004, what horse power is lost due to friction? 

164. Absorption Dynamometer. — The friction brake shown 
in Fig. 181 is used to absorb the energy of the mechanism. 
It may be used as a means of measuring the energy, and 
when so used it may be called an absorption dynamometer. 
The weight TF, attached to one end of the friction band, 
corresponds to the tension in the tight side of an ordinary 
belt (see Art. 156), while the force measured by the spring 
S corresponds to the tension on slack side of a belt. Let 
W= T^ and S = T^\ then 2\ = 7^2^-^% just as was found in 



306 APPLIED MECHANICS FOB ENGINEERS 

the case of belt tension. The work absorbed per revolu- 
tion is work =z (T^— T^}2 7rr^, where r^ is the radius of 
the brake wheel. The horse power absorbed is 

HP ^ (T,-T,r^^r, 

33,000 vv^>^\ . .^..v^ ^ ^ 

In many cases the friction band is a hemp rope, and in 
such cases it is possible to wrap the rope one or more 
times around the pulley, making it possible to make 
T^ — T^ large while T^ is small. 

The surface of the brake wheel may be kept cool by 
allowing water to flow over the inside surface of the rim, 
which should be provided with inside flanges for that 
purpose. 

165. Friction Brake. — The friction brake shown in 
Fig. 185 consists of the lever EQ^ the friction band, and 
the friction wheel. Such brakes are used on many types 
of hoisting drums, automobiles, etc. Let the band ten- 
sions be T^ and T^^ and let W be the force causing the 
motion, that is, the working force, and P the force applied 
at the end of the lever EC in such a way as to retard the 
rotation of the drum. We have here as before ^i = T^e^°- 
and the work per revolution C^i~ ^2)^ '^^1 + ^^ '^^s- 
By taking moments about A we have T^ — T^ = — 2 a, 

where r^ is the radius of the shaft and F is the force of 
friction acting on the shaft. Taking moments about (7, 

we have p^(^EC^ = T^d^ sin S + T^d^ sin /8. 

Problem 259. A weight of one ton is being lowered into a mine 
by means of a friction brake. The radius of the drum is 1^ ft., 



FRICTION 



307 



radius of the friction wlieel 2 ft., coefficient of brake friction .30, 
8 = 45^ ^ = 15^ r/j = cl^ = l ft, EC = G ft., radius of shaft 1 in., 
coefficient of axle friction .04, and the weight of the drum and brake 
wheel is 600 lb. Find T^, T,^, and P in order that the weight W 
may be lowered with uniform velocity. 

Problem 260. Suppose the weight in the above problem is being 
lowered with a velocity of 10 ft. per second when it is discovered that 
the velocity must be reduced one half while it is being lowered the 
next 10 ft., what pressure P will it be necessary to apply to the lever 
at E to make the change ? What will be the tension in the friction 
bands and the tension in the rope that supports TF? 

Problem 261. If the weight in the above problem has a velocity 
of 10 ft. per second, and it is required that the mechanism be so con- 
structed that it could be stopped in a distance of 6 ft., what pressure 
P on the lever and tensions T^ and T^g would it require ? What 
would be the tension in the rope caused by the sudden stop ? Com- 
pare this tension with IF, the tension when the motion is uniform. 

166. Prony Friction Brake. — The Prony friction brake 
may be used as an absorption dynamometer as shown in 




Fig. 194 



principle in Fig. 194. Let W be a working force acting 
on the wheel of radius r and suppose the brake wheel to 



308 APPLIED MECHANICS FOR ENGINEERS 

be of radius r^ The brake consists of a series of blocks 
of wood attached to the inner side of a metal band in 
such a way that it may be tightened around the brake 
wheel as desired by a screw at B, This band is kept 
from turning by a lever (7AZ>, held in the position shown 
by an upward pressure P, at A, Considering the forces 
acting on the brake and taking moments about the center, 
we have the couple due to friction, jFV^, equal to the mo- 
ment FQOA), or 

Fr^ = F(OA}. 

Considering the forces acting on the wheel, and neglecting 

axle friction, we get 

Fr-^ = Wr, 

The energy absorbed is used in heating the brake wheel. 
The wheel is kept cool by water on the inside of the rim. 
The work absorbed is 2 7rr^Fn= 2irF( 0A^7i^ where n is 
the number of revolutions. The force F may be measured 
by allowing a projection of the arm at A to press upon a 
platform scales. The horse power absorbed is 

where OA is expressed in feet, and n is the number of 
revolutions per minute. 

33 

If OAhQ taken as - — , a convenient length, the formula 

2 TT 

reduces to 

IT P -_^ 

A dynamometer slightly different from the Prony 
dynamometer is shown in Fig. 195. It differs only in the 
means of measurincr F. In this case the force F is meas- 



FRICTION 



309 



ured by the angular displacement of a lieavy pendulum 
Wy Taking moments about the axis of W^ and calling 




Fig. 195 

r^ the distance from that axis to its center of gravity and 
yS the angular displacement, we have 

Fr^ = W^Tr^ sin ^8, 

so that the horse power absorbed may be written 

7-4 33,000 ' 

where OA^ r^, and r^ are expressed in feet. If OA be 

taken as — ^, this becomes 



27r 



H.P.= 



W^r^ sin Pn 
r^lOOO 



The student should understand that the rotation of the 
mechanism at is not in every case due to a weight W 
being acted upon by gravity. In fact, in most cases, tlie 
motion will be due to the action of some kind of engine. 
This, however, will not change the expressions for horse 
power. 



310 APPLIED MECHANICS FOB ENGINEERS 

Problem 262. If ]]\ = 100 lb., OA = — , r^ = 2 ft., and r^ = 6 

27r 

in., what horse power is absorbed by the brake if /3 is 30°, and n is 

300 revolutions per minute ? 

167. Friction of Brake Shoes. — The application of the 
brake shoe to the wheel of an ordinary railway car is 
shown in Fig. 196, where F' is the axle friction, F the 
brake-shoe friction, N the normal pressure of the brake 

shoe, Gr the weight on 
the axle, and F^ and 
iV^j the reaction of the 
rail on the wheel. 
The brakes on a rail- 
way car when applied 
should be capable of 
absorbing all the en- 
ergy of the car in a 
very short time. The 
high speeds of modern 
trains require a system of perfectly working brakes, 
capable of stopping the car when running at its maximum 
speed in a very short distance. 

The coefficient of friction between the shoes and wheel 
for cast-iron wheels at a speed of 40 mi. per hour is about 
;|, while at a point 15 ft. from stopping the coefficient of 
friction is increased 7 per cent, or it is about .27. The 
coefficient for steel-tired wheels at a speed of 65 mi. per 
hour is .15, and at a point 15 ft. from stopping it is .10. 
(See Proc. M. C. B. Assoc, Vol. 39, 1905, p. 431.) 

The brake shoes act most efficiently when the force of 
friction F is as large as it can be made without causing a 




Fig. 196 



FRICTION 311 

slipping of the wheel on the rail (skidding). The normal 
pressure iV, corresponding to the values of the coefficient 
of friction given above, varies in brake-shoe tests from 
2800 lb. to 6800 lb., sometimes being as high as 10,000 
lb. 

Problem 263. A 20-ton car moving on a level track with a 
velocity of a mile a minute is subjected to a normal brake-shoe pres- 
sure of 6000 lb. on each of the 8 wheels. If the coefficient of brake 
friction is .15, how far will the car move before coming to rest? 

Problem 264. In the above problem the kinetic energy of rota- 
tion of the wheels, the axle friction, and the rolling friction have been 
neglected. The coefficient of friction for the journals is .002, that 
for rolling friction is .02. Each pair of wheels and axle has a mass 
of 45 and a moment of inertia with respect to the axis of rotation of 
37. The diameter of the wheels is 32 in. and the radius of the axles 
is 2i in. Compute the distance the car in the preceding problem will 
go before coming to rest. Compare the results. 

Problem 265. A '30-ton car is running at the rate of 70 mi. per 
hour on a level track when the power is turned off and brakes ap- 
plied so that the wheels are just about to slip on the rails. If the 
coefficient of friction of sliding between w^heels and rails is .20, how 
far will the car go before coming to rest? 

Problem 266. A 75-ton locomotive going at the rate of 50 mi. 
per hour is to be stopped by brake friction within 2000 ft. If the 
coefficient of friction is .25, what must be the normal brake-shoe 
pressure ? 

Problem 267. A 75-ton locomotive has its entire weight carried 
by five pairs of drivers (radius 3 ft.). The mass of one pair of drivers 
is 271 and the moment of inertia is 1830. If, when moving with a 
velocity of 50 mi. per hour, brakes are' applied so that slipping on 
the rails is impending, how far will it go before being stopped? The 
coefficient of friction between the wheels and rails is .20. 



312 APPLIED MECHANICS FOR ENGINEERS 

168. Train Resistance. — The resistance offered by a train 
depends upon a number of conditions, such as velocity, 
acceleration, the condition of track, number of cars, curves, 
resistance of the air, and grades. No law of resistance 
can be worked out from a theoretical consideration, be- 
cause of the uncertainty of the influence of the various 
factors involved. Formulae have been developed from the 
results of tests; the most important of these are given 
below. 

Let R represent the resistance in pounds and v the ve- 
locity in miles per hour. W. F. M. Goss has found that 
the resistance may be expressed as 

i2 = . 0003(^4- 347) ^;^ 

where L is the length of the train in feet (see Engineering 
Record, May 25, 1907). 

The Baldwin Locomotive Works have derived the 
formula 

b 

as the relation between the resistance and velocity. When 
all factors are considered, this becomes 

R = Z + ^+ .3788 (0 + .5682 {c) + .01265 (a)2, 

where t = grade in feet per mile, c the degree of curvature 
of the track, and a the rate of increase of speed in miles 
per hour in a run of one mile. 

To get the total resistance it is necessary to include, in 
addition to the above factors, tlie friction of the locomo- 



FRICTION 



313 



tive and tender. Tliis is given by Holmes (see Kent's 
"Hand Book") as 

R^= [12 + .3(i;- 10) W], 

where W is the weight of the engine and tender in pounds 
and R^ the resistance in pounds due to friction. 



70 



1 60 
oc 

LU 

°- 50 

CO 

o 

z 

2 40 



^ 30 



U^^W-^U::^!!-: ! .ltUl}!!4i^:-4!!.l- -^ :|.---!---H!.!-l U, -1; 



n FOR RAILROAD TRAINS 



-"d^ 




10 20 30 40 oO 60 70 bO 90 100 

VELOCITY IN MILES PER HOUR 

Fig. 197 

Other formulae derived as the result of experiments are 
shown graphically in Fig. 197. 



314 



APPLIED MECHANICS FOR ENGINEERS 



The formulae themselves are as follows (see Engineering, 
July 26, 1907): 



Curve Number 


Formula 


Authority 


1 


''='-^+;44 


Clark 


2 


'== « + m 


Clark 


3 


*=*-"+i';2 


Wellington 


4 


«=»+2» 


Deeley 


5 


i? = .2497 V 


Laboriette 


6 


R = 3.36 + .1867 v 


Baldwin Company 


7 


72 = 4.48 + 284:1; 


Lundie 


8 


J? = 2 + .24 i; 


Sinclair 


9 


72 = 2.5+ ""' 

65.82 


Aspinal 



It is evident that these formulae do not agree as closely 
as one would wish. The difference must be due chiefly to 
the different conditions under which the tests were made. 
These conditions should be taken into account in any 
application of the formulae to special cases. 



CHAPTER XV 

IMPACT 

169. Definitions. — ^When two bodies that are approach- 
ing each other collide, they are said to be subjected to 
impact. If their motion is along the line joining their 
centers of gravity, the collision is designated as direct 
central impact. If they are moving along parallel lines, 
not the common gravity line, the impact is known as direct 
eccentric impact. When the collision differs from either 
of the above forms, the impact is known as oblique impact. 

The phenomena of impact may be best studied by con- 
sidering the two bodies somewhat elastic. Suppose for 
simplicity that they are two spheres, M^ and M^^ Fig. 198, 
and that they are moving in opposite directions with 
velocities v^ and v^ and that the impact is central. In 
Fig. 198 (a) they are shown at the instant when contact 
first takes place, and in Fig. 198 (J) they are shown some 
time after first contact when each has been deformed 
somewhat by the pressure of the other. The dotted lines 
indicate the original spherical form and the full lines the 
actual form of the deformed spheres. When the spheres 
first touch, the pressure P between them is zero, but as 
each one compresses the other, the pressure P increases 
until it becomes a maximum. The compression of the 
spheres is indicated in the figure by d^ and d^. We shall 



316 



APPLIED MECHANICS FOR ENGINEERS 



designate the time during which the bodies are being com- 
pressed as the period of deformation. 

After the compression has reached its maximum value 
the bodies, if they be partially elastic, begin to separate and 

to regain their original 

_!j ^ form. The common 

pressure P decreases 
and becomes zero, if the 
bodies are sufficiently 
elastic so that they 
finally separate. We 
shall designate this pe- 
riod of separation as the 
period of restitution^ and 
the velocities of separa- 
tion as v-^^ and v^^ 

If the bodies are en- 
tirely inelastic^ there will 
be no restitution. They 
will, in that case, remain 
in contact just as they 
are when the pressure be- 
tween them is a maximum 
and will move on with a 
common velocity V. 




(a) 




170. Direct Central Impact, Inelastic. — When the bodies 
meet in direct central impact, separation will take place 
along the line joining tlie centers of gravity. Let T be 
the time from the first contact up to the time of maximum 
pressure, that is, the time of deformation^ and T^ the time 



IMPACT 317 

from first contact up to the time of separation. Then 
T^— T represents the time of restitution. We have, from 
Art. 72, dv = a ' dt. Considering the motion of M^ 
during the period of deformation, we have dv^ = a^dt 

anda,= -| 

so that I dv. = — -— : j Pdt^ 

or iffiC r - t^i) = - fpdt. 

Jo 

In a similar way, remembering that if v^ is positive v^ is 
negative, 

Jo 

The two integrals f Pdt on the right-hand side of the 

preceding equations cannot be determined since we do 
not know in general how the pressure P varies with the 
time; we do know, however, that they are equal term for 
term, so that we may eliminate them. We have, then, 



or 

If the bodies were moving in the direction of Jfj and 
Vj > v^, we should have both Vi and v^^ positive, a^ negative, 
and ^2 positive. Then 

M^ v^ + M^v^ 

This is also the value for P^if both bodies are moving in 
the direction v^ and v^ > v^. 



318 APPLIED MECHANICS FOR ENGINEERS 

If the bodies are inelastic, they will both move with the 
velocity V^ and there will be no separation. Suppose the 
bodies to be two lead balls, and let (7^ = 10 lb., Gr^ = 25 
lb., Vi = 10 ft. per second, and v^ = 60 ft. per second. 
Then V = 51.28 ft. per second if the bodies are moving in 
the same direction, and F^= 33.3 ft. per second if they 
move in opposite directions. 

The energy lost in direct central impact of inelastic 
bodies may be found by subtracting the kinetic energy of 
31-^ and 711^2, when the common velocity is F^ from the 
kinetic energy of the two bodies at the time of first con- 
tact. The kinetic energy of M^ before impact is E^ = 
I M^vl, and that of M^ is -£"2=2 ^2^% ^^ ^^^^^ ^^^ total 
energy before impact is ^M^v\ + ^M^v\, The kinetic 
energy after impact is ^(^M^ + M^ F"^, so that the loss of 
kinetic energy is 

and this equals ""Xlt^Mf 

if the bodies move in the same direction, or 

if they move in opposite directions. 

This energy is used up in deforming the bodies and in 
raising their temperatures. The kinetic energy remain- 
ing may be written 

when the bodies move in the same direction. 



IMPACT 319 

In the above example of the lead balls, we find that the 
kinetic energy lost due to impact is 40.3 ft. -lb. when the 
bodies move in the same direction, and 1109 ft. -lb. when 
they move in opposite directions. 

When the bodies move toward each other and M-i^i\ = 
M^v^^ the final velocity Vis zero and the kinetic energy lost 

is ^ M^v'^ + 1^ M^v'^, If the masses are equal, V= ^^ ~ ^2 
and the kinetic energy lost is 

MOh + v^ 

2 2 

If M^ is infinite as compared with 3f^^ and v^ is zero, the 
final velocity V is zero, and the kinetic energy lost is 

2 

Problem 268. A lead sphere whose radius is 2 in. strikes a large 

mass of cast iron after falling freely from rest through a distance 

of 100 ft. What is its final velocity? What is the loss of kinetic 
energy? 

Problem 269. A 10-lb. lead sphere is at rest when it is acted upon 
by another lead sphere, whose radius is 3 in., in direct central impact. 
The velocity of the latter sphere is 20 ft. per second. What is the 
common velocity of the two spheres and what is the loss of kinetic 
energy due to impact ? 

171. Direct Central Impact, Elastic. — If the impact is 
not too severe, elastic or partially elastic bodies tend to 
regain their original shape after the deformation has 
reached a maximum and finally separate if they possess 
sufficient elasticity. Using the notation of Art. 169, we 
have, for the period of restitution, if jR is the force of 
restitution. 



320 APPLIED MECHANICS FOR ENGINEERS 

M^\ dv^ = — J Rdt, and for M^, 

so that Mi(v[ - V) = -JRdt, 

and M^(v'^ - F) = Clidt 

The value of the integral j Pdt during deformation will 
not in general be the same as its value during restitution. 
Call the ratio of I Bdt to I Pdt, e. This value, which, is 

called the coefficient of restitution, is constant for a given 
material. It is unity for perfectly elastic substances, zero 
for non-elastic substances, and some intermediate value 
for the imperfectly elastic materials with which the 
engineer is usually concerned. The following values of e 
have been determined: for steel, ^ = .55, for cast iron, 
^ = 1, nearly; for wood, ^ = 0, nearly. 

From the above definition of e, it is seen at once that 
we may write 

v^ -V v^-V 
e-i = —, , and Co = ~ ' 



V- v{ 2 Y_ ^ 



and these equations enable one to determine e experi- 
mentally. 

Rdt = e ) Pdt, 

we may write 



IMPACT 321 

so that v[ — V{1 + gj) — e-^v^, 

and ^2 = V(l + ^2) - ^2^2' 

where ^=-E^r^' 

if the bodies are moving in the same direction, and 

M^ + M^ 

if they are moving in opposite directions. If the bodies 
are of the same material, e-^ = e^ = e. Then from the 
above equation it is seen that 

and the energy lost in impact is 

If the bodies are perfectly elastic, so that ^ = 1, the loss 
of energy is zero. 

Problem 270. The student should show that for any impact 

M^vi + M2V2 = M^v[ + il/g?;^ ; 

that is, the sum of the momenta before impact equals the sum of the 
momenta after impact. 

Problem 271. Two perfectly elastic bodies, having equal veloci- 
ties in opposite directions, meet in direct central impact. What must 
be the relation of their masses so that they will be reduced to rest? 

Problem 272. If the bodies are perfectly elastic and M^= M^, 
show that v[ = v^ and vl^ — Vy 

T 



322 APPLIED MECHANICS FOR ENGINEERS 

Problem 273. If a ball M^ of a certain material falls upon a 
large mass M2 of the same material from a height h and rebounds to 

a height /?,, show that e =^-^. 

^ h 

In this case il/g = 00 , rg = 0, and V = 0. 

Problem 274. A 20-ton car having a velocity of 40 mi. per hour 

collides with a 30-ton car having a velocity of 60 mi. per hour in the 
opposite direction. Both are destroyed. What is the loss of kinetic 
energy ? 

172. Elasticity of Material. — All materials of engineer- 
ing are imperfectly elastic. Some, however, show almost 
perfect elasticity for stresses that are rather low. This 
has been expressed by saying that all materials have a 
limit (elastic limit) beyond which if the stress be increased 
the material will be imperfectly elastic. Within the limit 
of elasticity^ stress is proportional to the deformation pro- 
duced. Let the total stress in tension or compression be 
P, and the stress, in pounds per square inch of cross sec- 
tion, be/, also let d be the deformation caused by JP and 
X the deformation per inch of length. Within the limit 

of elasticity of the material the ratio ^ is a constant, and 

A 

since /= — -, and X = -, when F is the area of cross sec- 
F I 

tion and I is the length of the material, it may be written 

Fl . . 

— — . This constant is called the modulus of elasticity of 
Ji d 

the material; it is usually represented by E^ so that 

for tension or compression. For steel F has been found to 
be 30,000,000 lb. per square inch. 



IMPACT 323 

173. Impact of Imperfectly Elastic Bodies. — Let d^ be 
the compression of M^ due to the impact, iincl c?2 the com- 
pression of ilfg* ^f -Pyji i^ tl^6 pressure between the two 
bodies when the compression is greatest, the average force 

p 

acting maybe represented by --^, if the limit of elasticity 

of the material is not passed. The work done on the two 

p 

bodies is -~ Qd^ + d.^^ and this should equal the energy 

lost during compression ; then 

p I 

From the preceding article we know that d. = ''L^ 

F I ' ' 

and c:?2= xT^^ where Zj and Zg are the lengths of the 

masses itf^ and M^ (considered prismatic), F^ and F^ the 
areas of cross section, and F^ and E^ the moduli of elas- 
ticity. The sum d^ + d^ may then be represented by 

^^J^l.^'l ^ J^2-^2 

F F F 7^ 

For convenience let ^ ^ = ^^ and ^ ^ = ^2 (iTj and 

^1 ^2 

^2 may be considered as representing the hardness). 
Then d, + d,==pj^j±fl 



and i>.. = (.^1- V,) ^3^^^^^^ l^r+ffJ 



324 APPLIED MECHANICS FOB ENGINEERS 

Problem 275. Suppose a 100-lb. steel hammer strikes an immov- 
able cast-iron plate with a velocity of 25 ft. per second. The hammer 
has a face of 3 sq. in. area and a length of 6 in. ; the plate has 
the same area and a thickness of 2 in. If the modulus of elas- 
ticity of steel is 30,000,000 and of cast iron is 15,000,000, find the 
greatest pressure between the two bodies due to the impact. 

Under the assumptions v^ = and Mo = oo , 

M^ = M., ij^ = 22,500,000, H^ = 15,000,000. 
o'2»2 

Then P^ = 132,750 lb. 

Problem 276. Let the mass of the ram of a pile driver be M^ 
and let h be the height of fall. Let il/^ be the mass of the pile and 
s its penetration under a blow. If c?^ is the compression of the ham- 
mer" and c?2 the compression of the pile, the work equation becomes 

Pr.s+^(d, + d,)= G,h, 

and P. = ,y , ■ 

It is required to find the load that the pile will carry. 
Since v^ = V2 gh and V2 = 0, we may write 



and then Pm = 



' ^ iV/i + M, V H,H^ J ' 

Gih 






8 + 



The load Pm that the pile will carry is found by measuring the 
penetration s for the last blow. It is customary to use a factor of 
safety, as was explained in Art. 137. 

Problem 277. A wooden pile whose cross section is 1.5 sq. ft., 
and whose length is 30 ft., is driven by a steel hammer of 2000 lb. 
weight falling a distance of 20 ft. The penetration at the last blow 
is observed to be \ in. What load will the pile carry, using a factor 
of safety of 6 ? 



IMPACT 325 

Assume the weight of the pile as 1800 lb., the modulus of elas- 
ticity of timber as 1,500,000, and of steel as 30,000,000, the face of 
the hammer 2 sq. ft., and its length 2 ft. 

An approximate formula may be obtained by noting that lo is 
large as compared with li and E^ is small as compared with Ei, so 
that Hi is large compared with Ho, thus making 

H. ^Ih Ih^ 1 . • ^ 1 N 

-Hjrr = -Hr=m (^Woximately). 

Substituting in the expression for Pm, we have 



■t^m. 



m 



_^ / MxM^gli 



(.Ml + M,)i/2 

Problem 278. The student should solve Problem 277, using this 
approximate formula and compare the result with that already 
obtained. 

Problem 279. Compare the results obtained in problems 277 
and 278 with those obtained by using formulas of Art. 137. 

174. Impact Tension and Impact Compression. — Figure 
199 (a) represents a massi!i2 subjected to impact from the 
mass M^ falling from rest through a height li. The mass 
M^ is compressed by the impact. Figure 199 (5) repre- 
sents the body M^ as subjected to impact in tension, the 
mass in this case being a rod having Gr^ attached to one end 
and the other end attached to a crosshead A, Tlie rod, 
crosshead, and weight fall freely together through the 
distance li until A strikes the stops at B. when one end 
of the rod suddenly comes to rest aud tlie weight Gr^ 
causes tension in the rod due to impact. 

Suppose v^ represents the velocity of M^ when impact 
occurs and l^ its length, whether it be a tension or compres- 



326 



APPLIED MECHANICS FOR ENGINEERS 



sion piece. Let V be the common velocity of the bodies 

at the time of greatest pressure. Then V= -^ ^ ^^ 

(see Art. 170), and the kinetic energy necessary to bring 
the two bodies to rest is given by the expression, 

where h is the height of fall. 




^1 



A 



B 



3/. 



M. 



G, 



{a) 



ib) 



Fig. 199 



This energy is used in stretching or compressing M2, if 
we neglect the work done on (7^, which is supposed small 
in comparison. Let c^g be the deformation of M^ when the 
pressure between the two bodies is greatest, and let the 



average pressure between them be -^, as before. Then the 



IMPACT 327 



work done on the bar may also be expressed as —^ x d^ 



2 

and P^d,=M^ = ^ML, 

2 ' 2^2 M^ + M^ 



where F^ is the cross section, l^ the length, and E^ the 
modulus of elasticity of M^. We may, therefore, write 



^-=\ir. 



Ml 2</LA 



+ if, E,F. 



2^ 2 



as the elongation or compression of M^ due to the impact. 

Problem 280. A weight of 500 lb. falls through a distance of 2 
ft. in such a way as to put a 1-in. round steel rod in tension. If 
the rod is 18 in. long, \yhat will be the elongation due to the im- 
pact ? 

In this case M, = ^^^ , M, = -M_, L = 18, h = 24, E. = 30,000,000, 

and F2 = j\ sq. in. 

Problem 281. A cylindrical piece of steel 1 in. high and 1 in. 
in diameter is subjected to compression by a weight of 20 lb. falling 
through a distance of 1 in. How much will it be compressed ? 

If we substitute for P^ its equivalent /2F2 (Art. 172), where /2 
represents the stress in il/g in pounds per square inch of cross section, 
we may write 

so that M^_J^^_^ 

Eo^ Ml + M2 E2F2 
which may be written 



2 E2 ' Ml + M2 

Since F'zl represents the volume of il/o, we have 

2 



h = l^ (vol.) 2 JZ!_+^^^- 



328 



APPLIED MECHANICS FOR ENGINEERS 



This equation represents the relation between the height of fall of 
Ml and the stress /2 produced by such fall. 

Problem 282. In the preceding problem what stress (pounds 
per square inch) was caused in the cylindrical block by the fall of 
the20-lb. weight? 

Problem 283. The safe stress in structural steel for moving loads, 
impact loads, is usually taken as 12,500 lb. per square inch (value of 
fo). Through what height might a 300-lb. weight fall so as to pro- 
duce tension in a 1-in. steel round bar, 10 ft. long, without ex- 
ceeding the safe stress? 

Problem 284. Two steel tension rods in a bridge, each of 2 
sq. in. in cross section and 20 ft. long, carry the effect of the 
impact of a loaded wagon as one wheel rolls over a stone 1 in. 
high. The weight on the wheel is 2000 lb. What stress is intro- 
duced in the tension rods ? 

Note. For the strength of metals under impact the student is 
referred to the work of W. K. Hatt, Am. Soc. " Testing Materials," 
Vol. lY, p. 282. 



175. Direct Eccentric Impact. — The impact is said to be 

direct eccentric when the 

line of motion does not 

31^ coincide with the line 

joining the centers of grav- 
ity of the two bodies (see 
Fig. 200) . Suppose M^ at 
rest and that it is acted 
upon by M2 moving with a 
velocity v^ in the direction 
shown, and that P,„ is the 
force exerted hy M^ on My 
We shall first show t^at the motion imparted to M^ may 



B 




> ? 












} 


4— 


-- 


'-. 


M, 






c 







Fig. 200 



IMPACT 329 

be considered a rotation about its center of gravity A com- 
bined with a translation of that center of gravity. 

Introduce at a point (<?), distant b below A^ two equal 

p 

and opposite forces equal to -^ and parallel to P^, The 

introduction of these forces does not change the state of 
motion of the body. Consider one half of P„ acting at B 
with I P^ acting at in the same direction. These two 
forces are equivalent to a single force F^ acting at A in 
the direction P^. The remaining ^ of P^ at B with the 
I P^ at (7 'form a couple, of moment PJ), which tends to 
produce rotation about the center of gravity A. The 
motion, then, imparted to 31^ may be considered as con- 
sistinor of a translation and a rotation. 

Considering the motion of 31^ and calling P the 
variable pressure between 31^ and 31^^ we may write 

3I^Cdv = Cpdt (see Art. 170), 

Ph 

and since rotation also occurs, and dco = 6dt = '^rrr^ ^^ 

(see Art. 103), we may also write 



k\3I^jJlco=hjFdt. 



These equations become upon integration 

3I^V=jPdt, 






b 

Considering now the motion of 3I2. we may write 



330 



APPLIED MECHANICS FOR ENGINEERS 



which gives 






Eliminating j Pdt from these equations, we have 



and 






and since F = w^ft, 

these equations are sufficient to determine Fand Wy 

Problem 285. If the bodies are both inelastic, find the kinetic 
energy lost in direct eccentric impact. 

Problem 286. Suppose M^ to be a bar of steel J in. in diameter 
and 2 ft. long, and suppose M2 to be a hammer weighing 2 lb. and 
that its velocity at the time of impact is 20 ft. per second, find V and 
Wj if the hammer strikes 10 in. from the center of the rod. 

Problem 287. Let M^ be a square stick of timber ^" x 4'' x 10' 
and let 1T/2 be a 10-lb. hammer having a velocity at the time of impact 
of 10 ft. per second. If the impact takes place 4 



B 



M, 



A 



I) 



P ft. from the center, find V and coj. 



176. Center of Percussion. 



We 



have 



seen that the motion of M^ in direct eccen- 
tric impact may be considered as being 
made up of a rotation combined with a 
translation. As a matter of fact, however, 
the motion that actually occurs is a rota- 
tion about an instantaneous center. We 
shall now find such center of rotation, 
called center of jjercussion (see Art. 106). 
Let M^, Fig. 201, be the body under 



Fig. 201 
consideration and let P be the force caused by the impact 



IMPACT 



331 



and h the distance of P from the center of rotation D. 
We have shown in the preceding article that 

h 
when Fand co^ are the velocities of the body at time of 
greatest pressure. Now V=(o^(li-h^, since D is mo- 
mentarily at rest. Therefore 



so that 






The quantities b and k^ are usually known, so that h may 
readily be computed. 

Problem 288. A right circular cone of steel, radius of whose 
base is 6 in. and altitude 6 in., is supported as a pendulum by an axis 
through its vertex parallel to the base. It is struck with a 3-lb. ham- 
mer with a velocity of 10 ft. per second, at the center of percussion. 
Find V and w at time of greatest pressure. 

Problem 289. A man strikes a blow with a steel rod 1^ in. in 
diameter and 4 ft. long, by holding the rod in the hand and striking 
the farther end against a stone in such away as to cause the rod to be 
under flexure. AVhere should he grasp the rod in order that he may 
receive no shock ? 

177. Oblique Impact of Body against Smooth Plane. — Let 
M (Fig. 202) be a sphere 
moving toward the plane 
indicated with a velocity 
at impact of t>, the direc- 
tion of motion making an 
angle a with the vertical 
to the plane. After im- Fig. 202 





J 


s 


© 


^^ 


^.^> 


'-V 


/ 


' ^ 


\/^ 


/• 






r^^^ 


/ 





332 



APPLIED MECHANICS FOB ENGINEERS 



pact the body iff rebounds with a velocity v-^ in the direc- 
tion, making an angle /3 with the vertical AB, Since 
the plane is considered smooth, the effect of the impact 
will be all in the direction of AB and the impulsive force 
after impact will be e times what it was before impact. 
Summing horizontal and vertical components of the 

velocities, we have 

v-^ sin /3 = V sin a ; 

v^ cos ^ = ev cos a. 
Dividing, we have 

tan yS = - tan a, 
e 

and squaring and adding. 

So that if a and e are known, v^ and yS may be determined. 
If the body is perfectly elastic, ^ == 1, /3 = «, and v-^ — v. 



If the body is inelastic, so that ^ = 0, /3 



TT 



9' "1 



V sm a, 



it then moves along the plane with a velocity v sin a. 

178. Impact of Rotating^ Bodies. — Suppose two bodies 
M^ ^^^d. 31^ revolve about two parallel axes and 0^ (Fig- 

203) in such a way 
that impact occurs at 
a point along the line 
BU, Let the point at 
which impact occurs be 
distant r^ from and 
rj from 0^ It is evi- 
dent that the kinetic 
energy oi M^ is I- Iq^I and that this is equal to the kinetic 




Fig. 203 



IMPACT 



333 



energy of an equivalent mass 3f situated at a distance r^ 

from O.suice IM^v- = l3f7'lco?^. Equating these values 

for kinetic energy, we have for the equivalent mass 31' 

AT J ^ 
the value— 2fo. Likewise the equivalent mass of 31^ at 

9 

71 T- 7 2 

the point of impact is —IJh, The impact of the two 

,2 

rotating bodies 31^ and iHf^ then, may be considered by 
considering the impact of their equivalent masses along 
the line DU, From Art. 171, we have 



and 



vl = F(l + e,) - e^v^ 

^2 = T^l + ^2) - ^2^'2' 



where V= 1^ 1 + 2^2 ^ if the bodies are moving in the 

same direction, and a similar expression with v^^ say, nega- 
tive, if they are moving in opposite directions. 

Let 0)2 be the angular velocity of 71^2 before impact. 

Let coi be the angular velocity of 31^ before impact. 

Let co^ be the angular velocity of 31^ after impact. 

Let col ^^ ^'^^ angular velocity of 31^ after impact. 
Then we may write 



co^r^ = t'2, (o^7\ = v^, co,^r^ = v.^, co[r^ = vl, 



so that 



G)i= r^ 



0)2 



C 3Lk^7'^ + 3J,k^i^~ 
■^ Oi 2 ^01 

3I,klrl + 3I,klrl 



2—1 



(1 + 0-^i«r 



(1 + ^2) -^2^2' 



334 APPLIED MECHANICS FOR ENGINEERS 

These equations may be put in the form 

1 2 2 1 



a>2 = (o^-r^ (rjft)! - r^to^yO- + e^} 



Ix 



1 2 ' 2 1 



Problem 290. Suppose the moment of inertia of M^ is 3000 and 
its angular velocity before impact one radian per second ; that of ilfg 




Fig. 204 



IMPACT 



335 



15,000 and its angular velocity zero. Let r^ = 2 ft. and rg = 3 ft. and 



= ^2 = 0. 

Then w^= .311 radian per second. 



0) 



'= — .207 radian per second. 



The kinetic energy lost due to the impact is 



1034 ft.-lb. 



Problem 291. A well drill is shown in principle in Fig. 204. 
The drill is supported by a cable that passes over a pulley C and is 
attached to a friction drum A. When A is held, the drill is raised by 
the operation of M^ and M^- Suppose that I is 300 and w^ = 3 
radians per second ; /g = 200 and w,^ = ] r^ = 2 ft. and rg = 6 ft. 
Assume e^ = e^^ J. Find o)[ and w'^. What kinetic energy is lost 
due to each impact ? 




Problem 292. The moment of inertia of the trip hammer ilfg, 
illustrated in principle in Fig. 205, is 100,000 ; that of J/j is 00,000. 
If Tj = 3 ft., r2 = 10 ft., (Oj = 2 radians per second, (Og = 0, and e^^ — e^ 
= i, find 0)1 and wL What is the kinetic energy lost due to each 
impact? What is the kinetic energy of the hammer? 



APPENDIX I 

HYPEEBOLIC FUNCTIONS 



cosh 



X 



sinh a; = 



tanh X = 



2 
sinh X e^ — e~^ 



cosh X e-^ + e ^ 



HYPERBOLIC FUNCTIONS 



339 



X 


Cosh X 


Sinha: 


X 


Cosh X 


Sinhx 


0.01 


1.0000,500 


0.0100002 


0.51 


1.13289:^ 


0.5323978 


.02 


.0002000 


.0200013 


.52 


.1382741 


.54375;^) 


.03 


.0004500 


.0300045 


.53 


.1437686 


.5551(;.37 


.04 


.0008000 


.0400107 


.54 


.1493776 


.5(;6()292 


.05 


.0012503 


.0500208 


.55 


.1551014 


.5781 51 (; 


.0() 


.0018006 


.06003()0 


.56 


.1609408 


.5897;n7 


.07 


.0024510 


.0700572 


.57 


.1668fK)2 


.()013708 


.08 


.0032017 


.0800854 


.58 


.1729()85 


.6130701 


.09 


.0040527 


.0901215 


.59 


.1791579 


.6248:)()5 


.10 


.0050042 


.1001668 


.60 


.1854652 


.636653(5 


.11 


.0060561 


.1102220 


.61 


.1918912 


.6485402 


.12 


.0072086 


.1202882 


.62 


.1984363 


.6604917 


.13 


.0084618 


.1303664 


.63 


.2051013 


.6725093 


.14 


.0098161 


.1404578 


.64 


.2118867 


.6845942 


.15 


.0112711 


.1505631 


.(>5 


.2187933 


.mmiry 


.1() 


.0128274 


.1606835 


.()() 


.2258219 


.7089704 


.17 


.0144849 


.1708200 


.67 


.2329730 


.7212()43 


.18 


.0162438 


.1809735 


.68 


.2402474 


.73:^)303 


.19 


.0181044 


.1911452 


.69 


.247()458 


.7460()97 


.20 


.0200668 


.2013360 


.70 


.2551690 


.7585837 


.21 


.0221311 


.2115469 


.71 


.2628178 


.7711735 


.22 


.0242977 


.22177i)0 


.72 


.2705927 


.7838405 


.23 


.02( )56( 58 


.2320333 


.73 


.2784948 


.7^5858 


.24 


.0289384 


.2423107 


.74 


.2865248 


.8094107 


.25 


.0314132 


.2526122 


.75 


.2946833 


.82231()7 


.26 


.0339<)08 


.2629393 


.76 


.3029713 


.835:>049 


.27 


.0366720 


.2732925 


.77 


.311385)6 


.84837()6 


.28 


.0394568 


.2836731 


.78 


.3199392 


.8()15330 


.29 


.0423456 


.2940819 


.79 


.3286206 


.8747758 


.30 


.0453385 


.3045203 


.80 


.3374349 


.8881060 


.31 


.0484361 


.3149891 


.81 


.3463831 


.9015249 


.32 


.0516384 


.3254894 


.82 


,3554658 


.9150342 


.33 


.05494()0 


.33()0222 


.83 


.3(>46840 


.928();U7 


.34 


.05835<)0 


.34(35886 


.84 


.3740388 


.9423282 


.35 


.0618778 


.3571898 


.85 


.3835309 


.9561 1(W 


.36 


.0(>55029 


.3(578265 


.86 


.3931614 


.9699993 


.37 


.0()92345 


.3785001 


.87 


.4029312 


.9839796) 


.38 


.07307:^ 


.3892116 


.88 


.4128413 


0.9980584 


.39 


.0770189 


.3999619 


.89 


.4228927 


l.()1223()9 


.40 


.0810724 


.4107523 


.90 


.4330864 


.02()5167 


.41 


.0852341 


.4215838 


.91 


.44.34234 


.0408991 


.42 


.0895042 


.4324574 


.92 


.4539048 


.055.3cS5() 


.43 


.0938888 


.4433742 


.93 


.4()45315 


.06)99777 


.44 


.0983718 


.4543354 


M 


.4753046 


.0<S4(i768 


.45 


.1029702 


.4()5;U20 


.95 


.4862254 


.0994 S43 


.46 


.1076788 


.4763952 


.9(> 


.4972947 


.1144018 


.47 


.1124983 


.4874959 


.97 


.50851.37 


.1294;i<)7 


.48 


.1174289 


.498()455 


.98 


.5198837 


.144:.72() 


.49 


.1224712 


.5098450 


.i)9 


.5314057 


.1598288 


0.50 


1.1276260 


0.5210953 


1.00 


1.54.30806 


1.1752012 



340 



HYPERBOLIC FUNCTIONS 



X 


Cosh X 


Sinli X 


X 


Cosh X 


Sinhx 


1.01 


1.5549100 


1.190(3910 


1.51 


.2338704 


2.1529104 


1.02 


.5668948 


.2062999 


1.52 


.3954(586 


.17675(36 


1.03 


.5790365 


.2220294 


1.53 


.41735()3 


.2008206 


l.Olr 


.5913358 


.2378812 


1.54 


.4394857 


.2251046 


1.05 


.6037945 


.2538567 


1.55 


.4618591 


.2496111 


1.0(3 


.6164134 


.2()99576 


1.56 


.4844787 


.2743426 


1.07 


.6291940 


.2861855 


1.57 


.5073467 


.2993014 


1.08 


.6421375 


.3025420 


1.58 


.5304654 


.3244903 


1.09 


.6552453 


.3190288 


1.59 


.5538373 


.3499117 


1.10 


.6685186 


.3356474 


1.60 


.5774645 


.3755679 


1.11 


.6819587 


.3523997 


1.61 


.()013494 


.4014618 


1.12 


.7005670 


.3(342872 


1.62 


.6254945 


.4275958 


1.13 


.7093449 


.3863116 


1.63 


.6499(^21 


.4539726 


1.14 


.7232938 


.4034746 


1.64 


.6745748 


.4805947 


1.15 


.7374148 


.4207781 


1.65 


.6995149 


.5074(350 


1.16 


.7517098 


.4382235 


1.66 


.7247249 


.5345859 


1.17 


.7661798 


.4558128 


1.67 


.7502074 


.5619()03 


1.18 


.7808265 


.4735477 


1.68 


.7759(550 


.5895910 


1.19 


.7956513 


.4914299 


1.69 


.8020001 


.6174806 


1.20 


.8106556 


.5094613 


1.70 


.8283154 


.6456319 


1.21 


.8258410 


.5276436 


1.71 


.8549136 


.6740479 


1.22 


.8412089 


.5459788 


1.72 


.8817974 


.7027311 


l!23 


.8567610 


.5644685 


1.73 


.9089692 


.7316847 


1.24 


.8724988 


.5831146 


1.74 


.9364319 


.7(509115 


1.25 


.8884239 


.(3019191 


1.75 


.9641884 


.7904143 


1.26 


.9045378 


.6208837 


1.76 


2.9922411 


.8201962 


1.27 


.9208421 


.6400105 


1.77 


3.0205932 


.8502601 


1.28 


.9373385 


.6593012 


1.78 


.0492473 


.8806091 


1.29 


.9540287 


.6787578 


1.79 


.0782063 


.9112461 


1.30 


.9709143 


.6983824 


1.80 


.1074732 


.9421742 


1.31 


1.9879969 


.7181768 


1.81 


.1370508 


2.9733966 


1.32 


2.0052783 


.7381431 


1.82 


.16()9421 


3.(W49163 


1.33 


.0227603 


.7582830 


1.83 


.1971501 


.03673(55 


1.34 


.0404446 


.7785989 


1.84 


.2276799 


.0(588(303 


1.35 


.0583329 


.7990926 


1.85 


.2585283 


.1012911 


1.36 


.0764271 


.8197662 


1.86 


.2897047 


.1340321 


1.37 


.0947288 


.840()219 


1.87 


.3212100 


.1670863 


1.38 


.1132401 


.8616615 


1.88 


.3530475 


.2004573 


1.39 


.1319627 


.8828874 


1.89 


.3852202 


.2341484 


1.40 


.1508985 


.9043015 


1.90 


.4177315 


.2681629 


1.41 


.1700494 


.9259060 


1.91 


.450584(3 


.3025041 


1.42 


.1894172 


.9477032 


1.92 


.4837827 


.3371758 


1.43 


.2090041 


.9(39()951 


1.93 


.5173293 


.3721810 


1.44 


.2288118 


1.9918840 


1.94 


.5512275 


.4075235 


1.45 


.2488424 


2.0142721 


1.95 


.5854808 


.44:52067 


1.46 


.2690979 


.0:>68616 


1.9() 


.6200927 


.4792343 


1.47 


.2895803 


.0591)549 


1.97 


.6550()(;7 


.515(5097 


1.48 


.3102917 


.082()540 


1.98 


.69040(31 


.55233(58 


1.49 


.3312341 


.1058614 


1.99 


.7261146 


.5894191 


1.50 


2.352409(3 


2.1292794 


2.00 


3.7()21957 


3.6268604 



HYPERBOLIC FUNCTIONS 



341 



X 


Cosh X 


Sinh X 


X 


Cosh X 


Siuh X 


2.01 


3.7986528 


3.6(54()642 


2.51 


6.19:!0993 


6.111S311 


2.02 


.8354899 


.7028:U5 


2.52 


.2545281 


.174()(5S5 


2.03 


.8727101 


.7413746 


2.53 


.31(55827 


.2:5692:57 


2.0-i 


.9103184 


.7802896 


2.54 


.3792(587 


.:3()(J402:3 


2.05 


.9483518 


.819(5198 


2.55 


.4425928 


.3(545111 


2.06 


.98()7111 


.8592571 


2.56 


.50(55(511 


.4292.~)(5:i 


2.07 


4.02550:'>8 


.8993179 


2.57 


.5711800 


.494(5444 


2.08 


.0()47395 


.9:598093 


2.58 


.(5:3(545(50 


.5(50(5820 


2.09 


.1043012 


.980(5140 


2.59 


.702:5958 


.(527:5758 


2.10 


.1443131 


4.0218567 


2.(50 


.7690059 


.6947:323 


2.11 


.1847398 


.06:^5018 


2.61 


.8362940 


.7627595 


2 12 


.2255846 


.1055530 


2.(52 


.9042(544 


.8314(515 


2.13 


.2668523 


.1480149 


2.63 


.9729254 


.<K1()84(59 


2.14 


.3085462 


.1908914 


2.64 


7.0422S;38 


.9709225 


2.15 


.3506713 


.2341871 


2.(55 


.112:34(33 


7.04169,10 


2.1(3 


..3932312 


.2779062 


2.(36 


.1831184 


.11:31701 


2.17 


.4362311 


.32205:34 


2.67 


.2546108 


.1853586 


2.18 


.4796741 


.366(5:325 


2.(58 


.32(58282 


.2582(550 


2.19 


.5235()49 


.411(5482 


2.69 


.3997785 


.3318975 


2.20 


.5679083 


.4571052 


2.70 


.4734686 


.40(52631 


2.21 


.6127086 


.5030079 


2.71 


.5479060 


.4813(392 


2.22 


.6579702 


.5493610 


2.72 


.62:30984 


.55722:37 


2.23 


.70:^1972 


.59(51688 


2.73 


.6990531 


.63:58338 


2.24: 


.7498951 


.(5434:564 


2.74 


.7757775 


.7112072 


2!25 


.71^)5677 


.6911(585 


2.75 


.85:32799 


.789:3520 


2.26 


.8437197 


.7:39:3(592 


2.76 


.9315(574 


.8(58275(5 


2.27 


.89135()5 


.7880444 


2.77 


8.010(5482 


.9479862 


2.28 


.9394824 


.8371982 


2.78 


.01K)5297 


8.0284911 


2.29 


.9881022 


.8868:358 


2.79 


.1712205 


.10<)7993 


2!30 


5.0372206 


.9369618 


2.80 


.2527285 


.1919185 


2.31 


.0868429 


.9875817 


2.81 


.3350617 


.27485(56 


2.32 


.1369741 


5.0:387004 


2.82 


.4182283 


.:358(5224 


2.33 


.1876186 


.0903228 


2.83 


.50223(58 


.41:52239 


2.^4 


.2387822 


.1424545 


2.84 


.587095(3 


.528(5(599 


2.35 


.2905196 


.1951504 


2.85 


.(57281:50 


.(5 1 45 H 587 


2.36 


.342()859 


.2482(556 


2.86 


.759:3979 


.7021291 


2.37 


.39543(55 


.3019558 


2.87 


.84(58585 


.7901595 


2.38 


.44872(56 


.35617(50 


2.88 


.9:552041 


.87<)0694 


2.39 


.5025(518 


.4109:321 


2.89 


9.02444:50 


.!i()SS(;()8 


2.40 


.5569472 


.4662293 


2.90 


.1145844 


9.0595611 


2.41 


.6118883 


.5220729 


2.91 


.2056373 


.1. -.11(316 


2.42 


.(5(573910 


.57.84(583 


2.92 


.2976105 


.24:5(57(59 


2.43 


.72:54594 


.(5:55422() 


2.93 


.:5905i:;8 


.: 5:571 1(58 


2.44 


.7801009 


.(5929401 


2.94 


.484:3.V,9 


.4314902 


2.45 


.837:»201 


.75102(55 


2.95 


.5791467 


.52(5S(^70 


2.4(3 


.8951232 


.809(5882 


2.9(3 


.(574S952 


.62:507(5:3 


2.47 


.9535159 


.8(5S9310 


2.97 


.771(5115 


.720:5()S1 


2.48 


6.012.'0:58 


.92S7(505 


2.98 


.8(59:5047 


.SIS.-) 119 


2.49 


.0720930 


.98918:51 


2.99 


.9(579S.")0 


.917(597(> 


2.50 


6.1322895 


6.0502045 


3.00 


10.0(37(3620 


10.017S750 



342 



n YPEliB OLIC FUN CTIONS 



X 


Coslia: 


Sinha: 


X 


Cosh X 


Sinha: 


3.01 


10.1683456 


10.1190539 


3.51 


16.7390823 


16.7091854 


3.02 


.2700464 


.2212451 


3.52 


.9070139 


.8774144 


3.03 


.3727741 


.3244585 


3.53 


17.0766361 


17.0473312 


3.04 


.4765391 


.4287042 


3.54 


.2479662 


.2189529 


3.05 


.5813518 


.5339929 


3.55 


.4210213 


.39229()6 


3.06 


.6872224 


.6403347 


3.56 


.5958178 


.567:3790 


3.07 


.7941620 


.7477408 


3.57 


.7723744 


.7442186 


3.08 


.9021809 


.85()2217 


3.58 


.9507082 


.9228:325 


3.09 


11.0112900 


.9657881 


3.59 


18.1308371 


18.1032388 


3.10 


.1215004 


11.0764511 


3.60 


.3127790 


.2854552 


3.11 


.2328226 


.1882217 


3.61 


.4965523 


.4695004 


3.12 


.3452684 


.3011112 


3.62 


.6821753 


.6553927 


3.13 


.4588488 


.4151309 


3.63 


.8696665 


.8431503 


3.14 


.5735748 


.5302919 


3.64 


19.05^)0447 


19.0327924 


3.15 


.6894584 


.6466062 


3.()5 


.2503288 


.2243376 


3.16 


.8065107 


.7640850 


3.66 


.4435377 


.4178052 


3.17 


.9247440 


.8827403 


3.67 


.6386909 


.61:32145 


3.18 


12.0441695 


12.0025838 


3.68 


.8358083 


.8105854 


3.19 


.1647998 


.123(5279 


3.69 


20.0349094 


20.0099373 


3.20 


.2866462 


.2458839 


3.70 


.2360140 


.2112905 


3.21 


.4097213 


.3693646 


3.71 


.4391421 


.4146645 


3.22 


.5340375 


.4940825 


3.72 


.6443142 


.6200802 


3.23 


.6596073 


.6200497 


3.73 


.8515505 


.8275577 


3.24 


.7864428 


.7472790 


3.74 


21.0608720 


21.0371178 


3.25 


.9145572 


.8757829 


3.75 


.2722997 


.2487819 


3.26 


13.0439629 


13.0055744 


3.76 


.4858548 


.4625710 


3.27 


.1746730 


.136{j665 


3.77 


.7015584 


.6785064 


3.28 


.30()7006 


.2690723 


3.78 


.9194324 


.89(J609() 


3.29 


.4400587 


.4028048 


3.79 


22.1394981 


22.1169025 


3.30 


.5747611 


.5378780 


3.80 


.3617777 


.3394069 


3.31 


.7108208 


.6743046 


3.81 


.5862933 


.5641452 


3.32 


.8482516 


.8120988 


3.82 


.8130681 


.7911403 


3.33 


.9870673 


.9512741 


3.83 


23.0421239 


23.0204143 


3.34 


14.1272820 


14.0918450 


3.84 


.2734843 


.2519907 


3.35 


.2689091 


.2338247 


3.85 


.5071715 


.4858917 


3.36 


.411 (KiO 


.3772277 


3.86 


.7432095 


.7221415 


3.37 


.5564583 


.5220686 


3.87 


.9816222 


.9607()38 


3.38 


.7024094 


.()(i83()19 


3.88 


24.2224327 


24.2017819 


3.39 


.8498306 


.8161219 


3.89 


.465()658 


.4452205 


3.40 


.9987366 


.9653634 


3.90 


.7113454 


.6911034 


3.41 


15.1491429 


15.1161016 


3.91 


.9594963 


.9394557 


3.42 


.3010()37 


.2683513 


3.92 


25.2101431 


25.1W3020 


3.43 


.4545147 


.4221278 


3.93 


.4633109 


.44:^6673 


3.44 


.6095114 


.5774468 


3.94 


.7190247 


.6995765 


3.45 


.7()60()88 


.7343232 


3.95 


.9773109 


.9580561 


3.46 


.9242033 


.8927735 


3.<HJ 


26.2:^.81943 


26.2191311 


3.47 


16.08:^)9298 


16.0528128 


3.97 


.5017019 


.4828285 


3.48 


.2452646 


.2144571 


3.98 


.7678597 


.7491740 


3.49 


.4082241 


.3777233 


3.99 


27.0:^()()943 


27.0181946 


3.50 


16.5728248 


16.5426275 


4.00 


.:5082:iU 


.2899175 



APPENDIX II 

LOGAKITHMS OF NUMBERS 



LOGARITHMS OF NUMBERS 



345 



LOGARITHMS OF NUMBERS, FROM TO 1000 


No. 





1 


2 


3 


4 


5 


6 


7 


8 


9 








00000 


30103 


47712 


6020(5 


69897 


77815 


84510 


90:309 


95424 


10 


00000 


00432 


008(50 


01283 


01703 


02118 


025:30 


02938 


03:342 


0:5742 


11 


04139 


04532 


04921 


05:307 


05(590 


0(50(59 


0(5445 


0(5818 


07188 


07554 


12 


07918 


08278 


08(536 


089<X) 


0i):i42 


09(591 


10037 


10:580 


10721 


1105<) 


13 


113^U 


11727 


12057 


12385 


12710 


13u:5:3 


i:5353 


1:3(572 


13987 


14:301 


14 


14613 


14921 


15228 


15533 


15836 


16136 


1(5435 


16731 


17026 


17:318 


15 


17609 


17897 


18184 


18469 


18752 


v.m's 


19:312 


195^)0 


198(55 


20139 


16 


20412 


20()82 


20951 


21218 


21484 


21748 


22010 


22271 


225:50 


22788 


17 


23045 


23299 


23552 


23804 


24054 


24303 


24551 


24797 


25042 


25285 


18 


25527 


25767 


2(5007 


2(5245 


2(5481 


2(5717 


26951 


27184 


27415 


27(546 


19 


27875 


28103 


28330 


28555 


28780 


29003 


29225 


29446 


296(56 


29885 


20 


30103 


30319 


30535 


30749 


30963 


31175 


31386 


31597 


31806 


:52014 


21 


3'79->2 


32428 


32(533 


32838 


33041 


33243 


3:3445 


33(546 


33845 


:34044 


22 


34242 


34439 


34(535 


34830 


35024 


:35218 


:35410 


35(502 


:35793 


:35983 


23 


36173 


363()1 


3(5548 


36735 


36921 


:^7106 


37291 


37474 


37657 


37839 


24 


38021 


38201 


38381 


38560 


38739 


38916 


39093 


39269 


39445 


39619 


25 


39794 


39967 


40140 


40312 


40483 


40654 


40824 


40993 


41162 


41330 


26 


41497 


416(i4 


41830 


41995 


421(>0 


42:524 


42488 


42651 


42813 


42975 


27 


43136 


43296 


4345(5 


4361(5 


43775 


4:3933 


44090 


44248 


44404 


44560 


28 


44716 


44870 


45024 


45178 


45331 


45484 


45(536 


45788 


459:3i) 


4(5089 


29 


46240 


46389 


46538 


46686 


46834 


46982 


47129 


47275 


47421 


47567 


30 


47712 


47856 


48000 


48144 


48287 


48430 


48572 


48713 


48855 


48995 


31 


49136 


49276 


49415 


49554 


49693 


49831 


49968 


50105 


50242 


50:379 


32 


50515 


50650 


50785 


50920 


51054 


51188 


5i:321 


51454 


51587 


51719 


33 


51851 


51982 


52113 


52244 


52374 


52504 


526:53 


52763 


52891 


53020 


34 


53148 


53275 


53402 


5352i) 


53655 


53781 


53907 


54033 


54157 


54282 


35 


54407 


54530 


54654 


54777 


54900 


55022 


55145 


55266 


55388 


55509 


36 


55630 


55750 


55870 


55990 


5(5110 


56229 


56:i48 


5(5466 


5(5584 


56702 


37 


56820 


56937 


57054 


57170 


57287 


57403 


57518 


57(5:34 


57749 


57863 


38 


57978 


58092 


5820(5 


58319 


58433 


5854(5 


58(558 


58771 


58883 


58995 


39 


59106 


59217 


59328 


59439 


59549 


59659 


597(59 


59879 


59988 


(50097 


40 


60206 


60314 


(50422 


60530 


60638 


60745 


60852 


60959 


61066 


61172 


41 


61278 


61384 


(51489 


61595 


61700 


61804 


61909 


(520i:5 


(52118 


(52221 


42 


62325 


(52428 


(52531 


62634 


(52736 


62838 


(52941 


(5:5042 


(53144 


(5:5245 


43 


63347 


();5447 


63548 


63(548 


63749 


6:3848 


(5:3948 


(54048 


64147 


(54246 


44 


64345 


64443 


64542 


(>4t540 


(54738 


(548:36 


649:53 


(550:30 


65127 


(55224 


45 


65321 


65417 


65513 


65()09 


(55705 


65801 


6589(5 


(55991 


(5(508(5 


(56181 


46 


66276 


(56370 


(5(5464 


(5(5558 


(5(5(551 


(5(5745 


6(58:58 


(5(1931 


(57024 


(57117 


47 


67210 


67302 


67394 


(57486 


(57577 


(57(5(59 


(57760 


(57851 


(57942 


(580:33 


48 


68124 


68214 


68:^4 


(58394 


(58484 


(58574 


(58(5(53 


(5S752 


(58842 


(589:50 


49 


69020 


69108 


69196 


(59284 


(59372 


(59460 


(59548 


(59(5: 55 


(59722 


(59810 


50 


69897 


69983 


70070 


70156 


70243 


70:^29 


70415 


70500 


7058(5 


70(571 


51 


70757 


70842 


70927 


71011 


7109(5 


71180 


712(55 


71:549 1 


714:53 


71516 


52 


71600 


71683 


717(57 


71850 


7193:5 


72015 


72098 


72181 


722(53 


72:545 


53 


72428 


72509 


72591 


72(572 


72754 


728:55 


72! 116 


72997 


7:5078 


73158 


54 


73239 


73319 


73399 


73480 


73559 


73639 


73719 


73798 


73878 


73957 



346 



LOGARITHMS OF NUMBERS 



LOGARITHMS OF NUMBERS, FROM TO 1000 




(Contl/tued) 


No. 





1 


2 


3 


4 


5 


6 


7 


8 9 


55 


74036 


74115 


74193 


74272 


74351 


74429 


74507 


74585 


74(5()3 


74741 


56 


74818 


74896 


74973 


75050 


75127 


75204 


75281 


' 75358 


75434 


75511 


57 


75587 


75663 


75739 


75815 


75891 


75966 


7()042 


76117 


76192 


76267 


58 


76342 


76417 


76492 


76566 


76641 


76715 


76789 


1 76863 


76937 


77011 


59 


77085 


77158 


77232 


77305 


77378 


77451 


77524 


77597 


77670 


77742 


60 


77815 


77887 


77959 


78031 


78103 


78175 


78247 


78318 


78390 


78461 


61 


78533 


78604 


78()75 


78746 


78816 


78887 


78958 


79028 


79098 


791(59 


62 


79239 


79309 


79379 


79448 


79518 


79588 


79()57 


79726 


79796 


798(55 


63 


79934 


80002 


80071 


80140 


80208 


80277 


80345 


80413 


80482 


80550 


64 


80618 


80685 


80753 


80821 


80888 


80956 


81023 


81090 


81157 


81224 


65 


81291 


81358 


81424 


81491 


81557 


81624 


81690 


81756 


81822 


81888 


6() 


81954 


82020 


82085 


82151 


82216 


82282 


82347 


82412 


82477 


82542 


67 


82607 


82672 


82736 


82801 


82866 


82930 


82994 


83058 


83123 


83187 


68 


83250 


83314 


83378 


83442 


83505 


83569 


83632 


83(595 


83758 


83821 


69 


83884 


83947 


84010 


84073 


84136 


84198 


84260 


84323 


84385 


84447 


70 


84509 


84571 


84633 


84695 


84757 


84818 


84880 


84941 


85003 


85064 


71 


85125 


85187 


85248 


85309 


85369 


85430 


85491 


85551 


85612 


85672 


72 


85733 


85793 


85853 


85913 


85973 


86033 


86093 


8(5153 


8(5213 


86272 


73 


86332 


86391 


86451 


86510 


86569 


86628 


86687 


86746 


86805 


86864 


74 


86923 


86981 


87040 


87098 


87157 


87215 


87273 


87332 


87390 


87448 


75 


87506 


87564 


87621 


87679 


87737 


87794 


87852 


87909 


87966 


88024 


76 


88081 


88138 


88195 


88252 


88309 


88366 


88422 


88479 


88536 


88592 


77 


88649 


88705 


88761 


88818 


88874 


88930 


88986 


89042 


8fK)98 


89153 


78 


89209 


89265 


89320 


89376 


89431 


89487 


89542 


89597 


89(552 


89707 


79 


89762 


89817 


89872 


89927 


89982 


90036 


90091 


90145 


90200 


90254 


80 


90309 


90363 


90417 


90471 


90525 


90579 


90633 


90687 


90741 


90794 


81 


90848 


90902 


90955 


91009 


91062 


91115 


91169 


91222 


91275 


91328 


82 


91381 


91434 


91487 


91540 


91592 


91645 


91698 


91750 


91803 


91855 


83 


91907 


91960 


92012 


92064 


92116 


92168 


92220 


92272 


92324 


92376 


84 


92427 


92479 


92531 


92582 


92634 


92685 


92737 


92788 


92839 


92890 


85 


92941 


92993 


93044 


93095 


93146 


93196 


93247 


93298 


93348 


93399 


86 


93449 


93500 


93550 


9;^01 


93651 


93701 


93751 


93802 


93852 


93^X)2 


87 


93951 


94001 


94051 


94101 


94151 


94200 


1M250 


94300 


94349 


94398 


88 


94448 


94497 


94546 


94596 


94645 


94(594 


94743 


94792 


94841 


94890 


89 


94939 


94987 


95036 


95085 


95133 


95182 


95230 


95279 


95327 


95376 


90 


95424 


95472 


95520 


95568 


95616 


95664 


95712 


95760 


95808 


95856 


91 


95904 


95951 


95999 


96047 


96094 


96142 


9(5189 


9(5236 


96284 


96331 


92 


%378 


96426 


9(^73 


9()520 


9()567 


9()614 


9(5661 


9(5708 


96754 


9(5801 


93 


96848 


96895 


9()941 


9()988 


970;U 


97081 


97127 


97174 


97220 


972(56 


94 


97312 


97359 


97405 


97451 


97497 


97543 


97589 


97635 


97680 


97726 


95 


97772 


97818 


97863 


97^K)9 


97954 


98000 


98045 


98091 


98136 


98181 


96 


98227 


98272 


98317 


983()2 


98407 


98452 


98497 


98542 


98587 


98632 


97 


98677 


98721 


9876(j 


98811 


98855 


98900 


98945 


98989 


99033 


99078 


98 


f)9122 


991()6 


99211 


99255 


99299 


99343 


99387 


99431 


99475 


99519 


99 


99563 


99607 


99651 


99694 


99738 


99782 


99825 


99869 


99913 


99956 



APPENDIX III 

TRIGONOMETRIC FUNCTIONS 



TRIGONOMETIilC FUNCTIONS 



349 



NATURAL SINES, COSINES, TANGENTS, ETC. 


O 


/ 


Sine 


Cosecant 


Tangent 


Cotangent 


Secant 


Cosine 


/ 











.000000 


Infinite 


.000000 


Infinite 


1.00000 


1.000000 





90 




10 


.002* 109 


343.77516 


.0(>29()9 


343.77371 


1.0()()(K) 


.99991 K3 


50 






20 


.oonsis 


171.8S831 


.005818 


171.88540 


1.00(X)2 


.999983 


40 






30 


.008727 


114.59301 


.008727 


114.588(55 


1.00004 


.999962 


30 






40 


.011()35 


85.945609 


.011()3() 


85.939791 


1.00007 


.999932 


20 






50 


.014544 


68.757360 


.014545 


68.750087 


1.00011 


.999894 


10 




1 





.017452 


57.298688 


.017455 


57.289962 


1.00015 


.999848 





89 




10 


.0203()1 


49.114062 


.020365 


49.103881 


1.00021 


.999793 


50 






20 


.0232()9 


42.975713 


.023275 


42.9(54077 


1.00027 


.999729 


40 






30 


.021)177 


38.201550 


.02()186 


38.188459 


i.ooo:^ 


.999657 


30 






40 


.029085 


34.382316 


.01^9097 


34.3(57771 


1.00042 


.999577 


20 






50 


.031992 


31.257577 


.032009 


31.241577 


1.00051 


.999488 


10 




2 





.034899 


28.653708 


.034921 


28.636253 


1.00061 


.999391 





88 




10 


.03780() 


2().450510 


.037834 


2(5.431(500 


1.00072 


.999285 


50 






20 


.040713 


24.562123 


.040747 


24.541758 


1.00083 


.999171 


40 






30 


.043(;i9 


22.925586 


.0436(51 


22.9037(36 


1.00095 


.999048 


30 






40 


.04(5525 


21.49:5676 


.046576 


21.470401 


1.00108 


.998917 


20 






50 


.049431 


20.230284 


.049491 


20.205553 


1.00122 


.998778 


10 




3 





.052336 


19.107323 


.052408 


19.0811.37 


1.00137 


.998630 





87 




10 


.055241 


18.102619 


.055325 


18.074977 


1.00153 


.998473 


50 






20 


.0.58145 


17.1984:i4 


.058243 


17.169337 


1.001(59 


.998:308 


40 






30 


.0()1049 


16.380408 


.0611(53 


16.:^9855 


1.00187 


.998135 


30 






40 


.C>63952 


15.6;i6793 


.0(54083 


15.(504784 


1.00205 


.997:357 


20 






50 


.006854 


14.957882 


.0(37004 


14.924417 


1.00224 


.997763 


10 




4 





.069756 


14.335587 


.069927 


14.300(366 


1.00244 


.997564 





86 




10 


.072658 


13.763115 


.072851 


13.72(5738 


1.002(35 


.997:357 


50 






20 


.075559 


13.234717 


.075776 


13.196888 


1.00287 


.997141 


40 






'SO 


.078459 


12.745495 


.078702 


12.70(3205 


1.00309 


.996917 


30 






40 


.081359 


12.291252 


.081629 


12.250505 


1.00333 


.99(3685 


20 






50 


.084258 


11.868370 


.084558 


11.826167 


1.00357 


.9iK3444 


10 




5 





.087156 


11.473713 


.087489 


11 430052 


1.00382 


.996195 





85 




10 


.090053 


11.104549 


.090421 


11.059431 


1.00408 


.9959:37 


50 






20 


.092950 


10.758488 


.093:^)4 


10.711913 


1.00435 


.995671 


40 






30 


.095846 


10.43;'.431 


.096289 


10.385397 


1.004(33 


.995396 


30 






40 


.098741 


10.127522 


.09922(5 


10.078031 


1.00491 


.995113 


20 






50 


.101635 


9.8391227 


.102164 


9.7881732 


1.00521 


.91M822 


10 




6 





.104528 


9.5667722 


.105104 


9.5143645 


1.00551 


.994522 





84 




10 


.107421 


9.3091()9<) 


.10804(5 


9.255;i035 


1.00582 


.994214 


50 






20 


.110313 


9.0651512 


.110990 


9.0098261 


1.00614 


.993897 


40 


83 


o 


/ 


Cosine 


Secant 


Cotangent 


Tangent 


Cosecant 


Sine 


f 





Fit func 


tions from S3* 


' 40' to 90° 


read from bott 


om of tab 


e upward. 



350 



TRIGONOMETRIC FUNCTIONS 



NATURAL SINES 


, COSINES, TANGENTS, ETC. 






(Co7itinued) 




o 


/ 


Sine 


Cosecant 


Tangent 


Cotangent 


Secant 


Cosine 


/ 


o 


6 


30 


.113203 


8.8336715 


.113936 


8.7768874 


1.00647 


.993572 


30 






40 


.11()093 


8.6137901 


.116883 


8.5555468 


1.00681 


.993238 


20 






50 


.118982 


8.4045586 


.119833 


8.3449558 


1.00715 


.992896 


10 




7 





.121869 


8.2055090 


.122785 


8.1443464 


1.00751 


.992546 





83 




10 


.124756 


8.0156450 


.125738 


7.9530224 


1.00787 


.992187 


50 






20 


.127642 


7.8344335 


.128694 


7.7703506 


1.00825 


.991820 


40 






30 


.130526 


7.6612976 


.131653 


7.5957541 


1.00863 


.991445 


30 






40 


.133410 


7.4957100 


.134613 


7.4287064 


1.00902 


.991061 


20 






50 


.136292 


7.3371909 


.137576 


7.2687255 


1.00942 


.990669 


10 




8 





.139173 


7.1852965 


.140541 


7.1153697 


1.00983 


.990268 





82 




10 


.142053 


7.0396220 


.143508 


6.9682335 


1.01024 


.989859 


50 






20 


.144932 


6.8997942 


.14()478 


6.8269437 


1.01067 


.989442 


40 






30 


.147809 


6.7654691 


.149451 


6.6911562 


1.01111 


.989016 


30 






40 


.150686 


6.6363293 


.152426 


6.5605538 


1.01155 


.988582 


20 






50 


.153561 


6.5120812 


.155404 


6.4348428 


1.01200 


.988139 


10 




9 





.156434 


6.3924532 


.158384 


6.3137515 


1.01247 


.987688 





81 




10 


.159307 


6.2771933 


.161368 


6.1970279 


1.01294 


.987229 


50 






20 


.162178 


6.1(360674 


.164354 


6.0844381 


1.01342 


.986762 


40 






30 


.165048 


6.0588980 


.167343 


5.9757644 


1.01391 


.986286 


30 






40 


.167916 


5.9553625 


.170334 


5.8708042 


1.01440 


.985801 


20 






50 


.170783 


5.8553921 


.173329 


5.7693688 


1.01491 


.985309 


10 




10 





.173648 


5.7587705 


.176327 


5.6712818 


1.01543 


.984808 





80 




10 


.176512 


5.6653331 


.179328 


5.5763786 


1.01595 


.984298 


50 






20 


.179375 


5.5749258 


.182332 


5.4845052 


1.01649 


.983781 


40 






30 


.182236 


5.4874043 


.185339 


5.3955172 


1.01703 


.983255 


30 






40 


.185095 


5.4026333 


.188359 


5.3092793 


1.01758 


.982721 


20 






50 


.187953 


5.3204860 


.191363 


5.2256647 


1.01815 


.982178 


10 




11 





.190809 


5.2408431 


.194380 


5.1445540 


1.01872 


.981627 





79 




10 


.193(i()4 


5.1635924 


.197401 


5.0658352 


1.01930 


.981068 


50 






20 


.196517 


5.0886284 


.200425 


4.9894027 


1.01989 


.980500 


40 






30 


.199368 


5.0158317 


.203452 


4.9151570 


1.02049 


.979925 


30 






40 


.202218 


4.9451687 


.206483 


4.8430045 


1.02110 


.979341 


20 






50 


.205065 


4.8764907 


.209518 


4.7728568 


1.02171 


.978748 


10 




12 





.207912 


4.8097343 


.212557 


4.7046e301 


1.02234 


.978148 





78 




10 


.210756 


4.7448206 


.215599 


4.6382457 


1.02298 


.977539 


50 






20 


.213599 


4.6816748 


.218645 


4.5736287 


1.02362 


.976921 


40 






30 


.21(>440 


4.()202263 


.221695 


4.5107085 


1.02428 


.97(>296 


30 






40 


.219279 


4.5604080 


.224748 


4.4494181 


1.02494 


.975(562 


20 






50 


.222116 


4.5021565 


.227806 


4.3896940 


1.02562 


.975020 


10 


77 


o 


f 


Cosine 


Secant 


Cotangent 


Tangent 


Cosecant 


Sine 


f 


o 


F 


or functions from 77° K 


' to S3° 30' read from bottom of ta 


ible upward. 



TRIGONOMETRIC FUNCTIONS 



351 



NATURAL SINES, COSINES, TANGENTS, ETC. 

(Continued) 



13 



14 



15 



16 



17 



18 



19 



r 


Sine 





.224951 


10 


.227784 


20 


.23061(3 


30 


.23;U45 


40 


.230273 


50 


.239098 





.241922 


10 


.244743 


20 


.247r)()3 


30 


.250380 


40 


.253195 


50 


.250008 





.258819 


10 


.201028 


20 


.2()44:34 


30 


.207238 


40 


.270040 


50 


.272840 





.275037 


10 


.278432 


20 


.281225 


30 


.284015 


40 


.280803 


50 


.289589 





.292372 


10 


.295152 


20 


.297930 


30 


.30070() 


40 


.30:3479 


50 


.300249 





.309017 


10 


.311782 


20 


.314545 


30 


.317305 


40 


.3200()2 


50 


.322810 





.325508 


10 


.328317 


20 


.331003 


1 


Cosine 



Cosecant 



4.4454115 
4.3<H)1158 
4.33()2150 
4.283«)57() 
4.2323i)43 
4.1823785 

4.1335055 
4.0859130 
4.0393804 
3.99392<)2 
3.9495224 
3.9001250 

3.8037033 
3.8222251 
3.7810590 
3.7419775 
3.7031.500 
3.0051518 

3.0279553 
3.59153()3 
3..5558710 
3.52093()5 
3.4807110 
3.4531735 

3.4203030 
3.3880820 
3.3504900 
3.3255095 
3.29512.34 
3.2053149 

3.2.300080 
3.2()73()73 
3.1791978 
3.151.54.")3 
3.124.3959 
3.09773)03 

3.07155.35 
3.0458.352 
3.0205093 



Serant 



Tangent 



.230808 
.233934 
.237004 
.240079 
.243158 
.240241 

.249328 
.252420 
.255517 
.2,58018 
.2()1723 
.204834 

.207949 
.271009 
.274195 
.277325 
.280400 
.283000 

.280745 
.28989(5 
.293052 
.2iH)214 
.299380 
.302553 

.305731 
.308914 
.312104 
.315299 
.318500 
.321707 

..324920 
.3281.39 
.331.304 
.3134.595 
.3.378.33 
.341077 

.344328 
.347585 
.350848 



Cotangent 



Cotangent 



4.3314759 
4.2747006 
4.2193318 
4.10.52998 
4.112.5014 
4.0010700 

4.0107809 
3.9010518 
3.9130420 
3.8007131 
3.8208281 
3.7759519 

3.7320508 
3.0890927 
3.04704()7 
3.()058835 
3.5055749 
3.5200938 

3.4874144 
3.4495120 
3.412302(5 
3.3759434 
3.340232(5 
3.3052091 

3.2708520 
3.2371438 
3.2040038 
3.1715948 
3.1397194 
3.1084210 

3.077()835 
3.0474915 
3.0178301 
2.98808.50 
2.9(500422 
2.9318885 

2.9042109 
2.8709970 
2.8502349 



Tangent 



Secant 


Cosine 


/ ' 


1.02(5.30 


.974370 





1.02700 


.973712 


50 


1.02770 


.973045 


40 


1.02842 


.972370 


30 


1.02914 


.971(587 


20 


1.02987 


.970995 


10 


1.0.3001 


.970296 





1.03137 


.9(59588 


50 


1.03213 


.<H58872 


40 


1.03290 


.9(58148 


30 


1.03.303 


.9(57415 


20 


1.03447 


.900075 


10 


1.0.3528 


.965926 





1.03(509 


.f)(551(59 


50 


1.03(591 


.9(544(J4 


40 


1.03774 


,9(5.3(5:50 


30 


1.03858 


.9(52849 


20 


1.03944 


.9(52059 


10 


1.04030 


.9(31262 





! 1.04117 


.900450 


50 


1 1.04200 


.9.59(542 


40 


1.04295 


.958820 


30 


1.04.385 


,957990 


20 


1.04477 


.957151 


10 


1.04509 


.950305 





1.04(5(33 


.955450 


50 


1.04757 


.954588 


40 


1.04853 


.95.3717 


30 


1.04950 


.9528:38 


20 


1.05047 


.951951 


10 


1.05146 


.951057 





1.0524(5 


.950154 


50 


1.05347 


.949243 


40 


1.0.5449 


.948:324 


30 


1.0.55.52 


.947:5!)7 


20 


1.05(5.57 


.94(5402 


10 


1.0.5762 


.945519 





1.058(59 


.<)44.508 


50 


1.0597(5 


.*^:3009 


40 


Cosecant 


Sine 


1 



77 



76 



75 



74 



73 



72 



71 
70 



For functions from 70° 40' to TT° 0' read from bottom of table upward. 



352 



TRIGONOMETRIC FUNCTIONS 



NATURAL 


SINES 


, COSINES, TANGENTS, ETC. 








(Cojitinued) 






O 


f 


Sine 


Cosecant 


Tangent 


Cotangent 


Secant 


Cosine 


/ 


o 


19 


30 


.333807 


2.9957443 


.354119 


2.8239129 


1.06085 


.942641 


30 






40 


.336547 


2.9713490 


.357396 


2.7980198 


1.0()195 


.941666 


20 






50 


.339285 


2.9473724 


.360680 


2.7725448 


1.06306 


.940684 


10 




20 





.342020 


2.9238044 


.363<)70 


2.7474774 


1.06418 


.939693 





70 




10 


.344752 


2.9006346 


.367268 


2.7228076 


1.06531 


.938694 


50 






20 


.347481 


2.8778532 


.370573 


2.6985254 


1.06645 


.937687 


40 






30 


.350207 


2.8554510 


.373885 


2.6746215 


1.06761 


.936672 


30 






40 


.352931 


2.8334185 


.377204 


2.6510867 


1.0()878 


.935650 


20 






50 


.355651 


2.8117471 


.380530 


2.6279121 


1.06995 


.934619 


10 




21 





.358368 


2.7904281 


.383864 


2.6050891 


1.07115 


.933580 





69 




10 


.3()1082 


2.7694532 


.387205 


2.5826094 


1.07235 


.932534 


50 






20 


.363793 


2.7488144 


.390554 


2.5604649 


1.07356 


.931480 


40 






30 


.366501 


2.7285038 


.393911 


2.5386479 


1.07479 


.930418 


30 






40 


.369206 


2.7085139 


.397275 


2.5171507 


1.07602 


.929348 


20 






50 


.371908 


2.6888374 


.400647 


2.4959661 


1.07727 


.928270 


10 




22 





.374607 


2.6694672 


.404026 


2.4750869 


1.07853 


.927184 





68 




10 


.377302 


2.()503962 


.407414 


2.4545061 


1.07981 


.926090 


50 






20 


.379994 


2.6316180 


.410810 


2.4342172 


1.08109 


.924989 


40 






30 


.382683 


2.6131259 


.414214 


2.4142136 


1.08239 


.923880 


30 






40 


.385369 


2.5949137 


.417626 


2.3944889 


1.08370 


.922762 


20 






50 


.388052 


2.5769753 


.421046 


2.3750372 


1.08503 


.921638 


10 




23 





.390731 


2.5593047 


.424475 


2.3558524 


1.08636 


.920505 





67 




10 


.393407 


2.5418961 


.427912 


2.3369287 


1.08771 


.919364 


50 






20 


.396080 


2.5247440 


.431358 


2.3182606 


1.08907 


.918216 


40 






30 


.398749 


2..5078428 


.434812 


2.2998425 


1.09044 


.917060 


30 






40 


.401415 


2.4911874 


.438276 


2.2816693 


1.09183 


.915896 


20 






50 


.404078 


2.4747726 


.441748 


2.2637357 


1.09323 


.914725 


10 




24 





.40()737 


2.4585933 


.445229 


2.2460368 


1.09464 


.913545 





66 




10 


.409392 


2.4426448 


.448719 


2.2285676 


1.09606 


.912358 


50 






20 


.412045 


2.4269222 


.452218 


2.2113234 


1.09750 


.911164 


40 






30 


.414()93 


2.4114210 


.455726 


2.1942997 


1.09895 


.909961 


30 






40 


.417338 


2.3961367 


.459244 


2.1774920 


1.10041 


.908751 


20 






50 


.419980 


2.3810650 


.462771 


2.1608958 


1.10189 


.907533 


10 




25 





.422618 


2.3662016 


.466308 


2.1445069 


1.10338 


.906308 





65 




10 


.425253 


2.3515424 


.4()9854 


2.1283213 


1.10488 


.VX)5075 


50 






20 


.427884 


2.3370833 


.473410 


2.1123348 


1.10640 


.903834 


40 






30 


.430511 


2.3228205 


.47()976 


2.096543() 


1.10793 


.902585 


30 






40 


.433135 


2.3087501 


.480551 


2.0809438 


1.10947 


.901329 


20 






50 


.435755 


2.2948685 


.484137 


2.0655318 


1.11103 


.900065 


10 


64 


o 


r 


Cosine 


Secant 


Cotangent 


Tangent 


Coserant 


Sine 


/ 


o 


I 


'or functioi 


IS from 64° -1 


y to 70° -30' read from b 


ottom of U 


ible upward. 



TRIGONOMETRIC FUNCTIONS 



353 



NATURAL SINES 


, COSINES, TANGENTS, ETC. 






(^Contuiued) 




O 


t 


Sine 


Cosecant 


Tangent 


Cotangent 


Secant 


Cosine 


/ 


o 


26 





.438371 


2.2811720 


.487733 


2.0503038 


1.11260 


.898794 





64 




10 


.440984 


2.2(57(5571 


.491339 


2.03525(55 


1.11419 


.897515 


50 






20 


.443593 


2.2543204 


.494955 


2.0203862 


1.11579 


.89(5229 


40 






30 


.44(5198 


2.2411585 


.498582 


2.0056897 


1.11740 


.8949154 


30 






40 


.448799 


2.2281(581 


.502219 


1.9911(537 


1.111K)3 


.893(533 


20 






50 


.451397 


2.2153460 


.505867 


1.9768050 


1.12067 


.892323 


10 




27 





.453990 


2.2026S93 


.509525 


1.9(526105 


1.12233 


.891007 





63 




10 


.45()580 


2.1901947 


.513195 


1.9485772 


1.12400 


.889(582 


50 






20 


.4591()<) 


2.1778595 


.51(5876 


1.9347020 


1.125(58 


.888350 


40 






30 


.4()1749 


2.1(55(5806 


.520567 


1.9209821 


1.12738 


.887011 


30 






40 


.4(54327 


2.153(5553 


.524270 


1.9074147 


1.12910 


.885(5(54 


20 






50 


.466901 


2.1417808 


.527984 


1.8939971 


1.13083 


.884309 


10 




28 





.4(59472 


2.1300545 


.531709 


1.8807265 


1.13257 


.882948 





62 




10 


.472038 


2.1184737 


.535547 


1.867(5003 


1.13433 


.881578 


50 






20 


.474(500 


2.1070359 


.539195 


1.8546159 


1.13(510 


.880201 


40 






30 


.477159 


2.0957385 


.542956 


1.8417409 


1.13789 


.878817 


30 






40 


.479713 


2.0845792 


.546728 


1.8290(528 


1.13970 


.877425 


20 






50 


.482263 


2.0735556 


.550515 


1.8164892 


1.14152 


.876026 


10 




29 





.484810 


2.0626653 


.554309 


1.8040478 


1.14335 


.874620 





61 




10 


.487352 


2.05190(51 


.558118 


1.79173(52 


1.14521 


.87320(5 


50 






20 


.489890 


2.0412757 


.561939 


1.7795524 


1.14707 


.871784 


40 






30 


.492424 


2.0307720 


.565773 


1.7(574940 


1.14896 


.87035(5 


30 






40 


.494953 


2.020:592i) 


.5(59619 


1.7555.590 


1.15085 


.8(58920 


20 






50 


.497479 


2.0101362 


.573478 


1.7437453 


1.15277 


.867476 


10 




30 





.500000 


2.0000000 


.577350 


1.7320508 


1.15470 


.8(56025 





60 




10 


.502517 


1.9899822 


.581235 


1.7204736 


1.156(55 


.8(545(57 


50 






20 


.505030 


1.9800810 


.5851;^ 


1.7090116 


1.158(51 


.8(53102 


40 






30 


.507538 


1.9702944 


.589045 


1.(597(5(531 


1.1(5059 


.8(51(529 


30 






40 


.510043 


1.960(5206 


.592970 


1.68(54261 


1.1(5259 


.8(50149 


20 






50 


.512543 


1.9510577 


.596908 


1.6752988 


1.16460 


.858662 


10 




31 





.515038 


1.9416040 


.600861 


1.6642795 


1.16663 


.857167 





59 




10 


.517529 


1.9:522578 


.604827 


1.6533(5(53 


1.1(58(58 


.855(5(55 


50 






20 


.520016 


1.9230173 


.608807 


1.(5425576 


1.17075 


.85415(5 


40 






30 


.522499 


1.9138809 


.612801 


1.(5318517 


1.17283 


.852(540 


30 






40 


.524977 


1.90484(5!» 


.61(5809 


1.(52124(59 


1.17493 


.851117 


20 






50 


.527450 


1.8959138 


.620832 


1.6107417 


1.17704 


.849586 


10 




32 





.529919 


1.8870799 


.624869 


1.6003345 


1.17918 


.848048 





58 




10 


.532384 


1.8783438 


.(528921 


1.5900238 


1.18133 


.84(i.")03 


50 






20 


.534844 


1.8697040 


.632988 


1.5798079 


1.18350 


.844951 


40 


57 


o 


r 


Cosine 


Serant 


Cotaiijrent 


Tangent 


Cosecant 


Sine 


f 


o 




For functions from 5T°" 


[(Y to 04°-0' read from bottom of ta 


ble upward. 



2 A 



354 



TRIGONOMETRIC FUNCTIONS 



6 



NATURAL 


SINES 


, COSINES, TANGENTS, ETC. 








(CoJitijiued) 






o 


I 


Sine 


Cosecant 


Tangent 


Cotangent 


Secant 


Cosine 


/ 


o 


32 


30 


.537300 


1.8611590 


.637079 


1.5696856 


1.18569 


.843391 


30 






40 


.539751 


1.8527073 


.641167 


1.5596552 


1.18790 


.841825 


20 






50 


.542197 


1.8443476 


.645280 


1.5497155 


1.19012 


.840251 


10 




33 





.544639 


1.8360785 


.649408 


1.5398650 


1.19236 


.838671 





57 




10 


.54707() 


1.8278985 


.653531 


1.5301025 


1.19463 


.837083 


50 






20 


.549509 


1.8198065 


.657710 


1.5204261 


1.19691 


.835488 


40 






30 


.551937 


1.8118010 


.661886 


1.5108352 


1.19920 


.833886 


30 






40 


.554360 


;L.8038809 


.666077 


1.5013282 


1.20152 


.832277 


20 






50 


.556779 


1.7960449 


.670285 


1.4919039 


1.20386 


830661 


10 




34 





.559193 


1.7882916 


.674509 


1.4825610 


1.20622 


.829038 





56 




10 


.561602 


1.7806201 


.678749 


1,4732983 


1.20859 


.827407 


50 






20 


.564007 


1.7730290 


.68:^.007 


1.4641147 


1.21099 


.825770 


40 






30 


.56<)406 


1.7655173 


.687281 


1.4550090 


1.21341 


.824126 


30 






40 


.568801 


1.7580837 


.691573 


1.4459801 


1.21584 


.822475 


20 






50 


.571191 


1.7507273 


.695881 


1.4370268 


1.21830 


.820817 


10 




35 





.573576 


1.7434468 


.700208 


1.4281480 


1.22077 


.819152 





55 




10 


.575957 


1.7362413 


.704552 


1.4193427 


1.22327 


.817480 


50 






20 


.578332 


1.7291096 


.708913 


1.4106098 


1.22579 


.815801 


40 






30 


.580703 


1.7220508 


.713293 


1.4019483 


1.22833 


.814116 


30 






40 


.583069 


1.7150639 


.717691 


1.3933571 


1.23089 


.812423 


20 






50 


.585429 


1.7081478 


.722108 


1.3848355 


1.23347 


.810723 


10 




36 





.587785 


1.7013016 


.726543 


1.3763810 


1.23607 


.809017 





54 




10 


.590136 


1.6945244 


.730996 


1.3679959 


1.23869 


.807304 


50 






20 


.592482 


1.6878151 


.735469 


1.3596764 


1.24134 


.805584 


40 






30 


.594823 


1.6811730 


.739961 


1.3514224 


1.24400 


.803857 


30 






40 


.597159 


1.6745970 


.744472 


1.3432331 


1.24669 


.802123 


20 






50 


.599489 


1.6680864 


.749003 


1.3351075 


1.24940 


.800383 


10 




37 





.601815 


1.6616401 


.753554 


1.3270448 


1.25214 


.798636 





53 




10 


.604136 


1.6552575 


.758125 


1.3190441 


1.25489 


.796882 


50 






20 


.606451 


1.6489376 


.762716 


1.3111046 


1.25767 


.795121 


40 






30 


.608761 


1.6426796 


.767627 


1.3032254 


1.26047 


.793353 


30 






40 


.611067 


1. (5364828 


.771959 


1.2954057 


1.26330 


.791579 


20 






50 


.613367 


1.6303462 


.776612 


1.2876447 


1.26615 


.789798 


10 




38 





.615661 


1.6242692 


.781286 


1.2799416 


1.26902 


.788011 





52 




10 


.617951 


1.6182510 


.785981 


1.2722957 


1.27191 


.786217 


50 






20 


.620235 


1.6122fK)8 


.790698 


1.2647062 


1.27483 


.784416 


40 






30 


.622515 


1.6063879 


.795436 


1.2571723 


1.27778 


.782608 


30 






40 


.624789 


1.6005416 


.800196 


1.2496933 


1.28075 


.780794 


20 






50 


.627057 


1.5947511 


.804080 


1.2422685 


1.28374 


.778973 


10 


51 


o 


1 


Cosine 


Secant 


Cotangent 

1 


Tangent 


Cosecant 


Sine 


/ 


o 


I 


"or functio 


ns from 51°-1 


0' to 57° -30' read from b 


ottom of tf 


ible iii)\var(l. 



TRIGONOMETRIC FUNCTIONS 



355 



NATURAL 


SINES 


>, COSINES, TANGENTS, ETC. 








(Continued) 




O 


/ 


Sine 


Cosecant 


Tangent 


Cotangent 


Secant 


Cosine 


/ 





39 





.629320 


1.5890157 


.809784 


1.2348972 


1.28676 


.77714(5 





51 




10 


.(531578 


1.5833318 


.814612 


1.2275786 


1.28980 


.775312 


50 






20 


.633831 


1.5777077 


.8194(53 


1.2203121 


1.29287 


.77:3472 


40 






30 


.636078 


1.5721337 


.82433(5 


1.21:30970 


1 .29597 


.771625 


30 






40 


.638320 


1.5666121 


.8292:U 


1.2059:527 


1.29909 


.7(59771 


20 






50 


.640557 


1.5611424 


.834155 


1.1988184 


1.30223 


.767911 


10 




40 





.642788 


1.5557238 


.839100 


1.1917536 


1.30541 


.766044 





50 




10 


.645013 


1.5503558 


.8440(59 


1.1847:576 


l.:50861 


.7(54171 


50 






20 


.647233 


1.5450378 


.849062 


1.1777698 


1.31183 


.7(52292 


40 






30 


.649448 


1.5397(590 


.854081 


1.1708496 


1.31509 


.7(50406 


30 






40 


.651657 


1.5345491 


.859124 


1.1(539763 


1.31837 


.758514 


20 






50 


.653861 


1.5293773 


.864193 


1.1571495 


1.32168 


.756615 


10 




41 





.656059 


1.5242531 


.869287 


1.1503684 


1.32501 


.754710 





49 




10 


.658252 


1.5191759 


.874407 


1.143(5326 


1.328:38 


.752798 


50 






20 


.6(50439 


1.5141452 


.879553 


1.1:369414 


1.33177 


.750880 


40 






30 


.(362620 


1.5091(505 


.884725 


1.1302944 


l.:3:3519 


.748956 


30 






40 


.()6479() 


1.5042211 


.889924 


1.1236909 


1.3:38(54 


.747025 


20 






50 


.666966 


1.4993267 


.895151 


1.1171:305 


1.34212 


.745088 


10 




42 





.669131 


1.4944765 


.900404 


1.1106125 


1.34563 


.743145 





48 




10 


.671289 


1.489(5703 


.W5(585 


1.1041:5(55 


1.34917 


.741195 


50 






20 


.67^43 


1.4849073 


.910994 


1.0977020 


l.:35274 


.739239 


40 






30 


.675590 


1.4801872 


.916331 


1.091:5085 


l.:356:54 


.737277 


30 






40 


.677732 


1.4755095 


.921(597 


1.0849554 


1.35997 


.735309 


20 






50 


.679868 


1.4708736 


.927091 


1.0786423 


1.36363 


.733335 


10 




43 





.681998 


1.4662792 


.932515 


1.0723(587 


1.36733 


.73ia54 





47 




10 


.684123 


1.4(517257 


.937iH)8 


i.066i:ui 


1.37105 


.7293(57 


50 






20 


.686242 


1.4572127 


.943451 


1.0599:381 


l.:37481 


.727374 


40 






30 


.688355 


1.4527397 


.9481K55 


1.0537801 


l.:378t50 


.725374 


30 






40 


.6904()2 


1.448:^(53 


.954508 


1.0476598 


1.38242 


.72336<) 


20 






50 


.692563 


1.4439120 


.960083 


1.0415767 


1.38628 


.721357 


10 




44 





.694658 


1.4395565 


.9(55689 


1.0355:503 


1.39016 


.719340 





46 




10 


.(i9()748 


1.4352393 


.97132(5 


1.0295203 


l.:594()9 


.717:316 


50 






20 


.()98832 


1.4:509()02 


.97(59i)(5 


1.02:354(51 


l.:39804 


.715286 


40 






30 


.700{K)9 


1.42(57182 


.982<5i>7 


1.01 76074 


1.4020:5 


.71:5251 


30 






40 


.702981 


1.4225134 


.988432 


1.0117088 


1.40(50(5 


.711209 


20 






50 


.705047 


1.4183454 


.994199 


1.0058:348 


1.41012 


.709161 


10 




45 





.707107 


1.4142136 


1.000000 


1.0000000 


1.41421 


.707107 





45 


o 


/ 


Cosine 


Secant 


Cotangent 


Tangent 


Cosecant 


Sine 


/ 







For functit 


)ns from 45°-( 


y to 51°-0' reiul from bottom of tab 


le upward. 



APPENDIX IV 

SQUARES, CUBES, SQUARE ROOTS, ETC. 



SQUARES, CUBES, SQUARE ROOTS, ETC. 



359 



SQUARES, CUBES 


, SQUARE ROOTS, 


CUBE 




BOOTS, AND RECIPROCALS 




No. 


Squares 


Cubes 


Square Roots 


Cube Roots 


Reciprocals 


1 


1 


1 


1.0000000 


1.0000000 


1.000000000 


2 


4 


8 


1.414213() 


1.2599210 


.500000000 


3 


9 


27 


1.7320508 


1.4422496 


.;3:3:5:5:3:5:3:53 


4 


16 


64 


2.0000000 


1.5874011 


.250000000 


5 


25 


125 


2.23(J0(i80 


1.7099759 


.200000000 


6 


36 


216 


2.4494897 


1.8171206 


.l(i(;6( 56(567 


7 


49 


343 


2.()457513 


1.9129312 


.142.S.57143 


8 


64 


512 


2.8284271 


2.00000(J0 


.] 25000000 


9 


81 


729 


3.0000000 


2.0800837 


.111111111 


10 


100 


1000 


3.1622777 


2.1544;M7 


.100000000 


11 


121 


1331 


3.316()248 


2.2239801 


.090iK)<M)01 


12 


144 


1728 


3.4()41016 


2.2894286 


.08:3:33:3:533 


13 


169 


2197 


3.6055513 


2.3513347 


.07(592:3077 


14 


196 


2744 


3.7416.574 


2.4101422 


.071428571 


15 


225 


3375 


3.87298:53 


2.4(i(52121 


.0(5(5(5(5(5(5(57 


16 


256 


4096 


4.0000000 


2.5198421 


.062.")0OO()0 


17 


289 


4913 


4.1231056 


2.5712816 


.(J58S2:3529 


18 


324 


5832 


4.2426407 


2.6207414 


.05." 55.").". "5(5 


19 


361 


6859 


4.3588989 


2.6684016 


.052(531579 


20 


400 


8000 


4.4721360 


2.7144177 


.050000000 


21 


441 


9261 


4.5825757 


2.7589243 


.047619048 


22 


484 


10648 


4.6904158 


2.8020:393 


.04.">454545 


23 


529 


12167 


4.7958315 


2.8438670 


.04:347S2(51 


24 


576 


13824 


4.8989795 


2.8844991 


.041(5(56(5(57 


25 


625 


15625 


5.0000000 


2.9240177 


.040000000 


26 


676 


17576 


5.091K)195 


2.9()24960 


.0:38461. "):38 


27 


729 


11M383 


5.1<H;1524 


3.0000000 


.0370370:37 


28 


784 


21952 


5.2i)15026 


3.0:^(>')889 


.0:357142S(5 


29 


841 


24389 


6.3851648 


3.07231(58 


.034482759 


30 


900 


27000 


5.4772256 


3.1072325 


.0»5:):).j>). 5:3.3 


31 


961 


29791 


5.5(577644 


3.1413806 


.0:52258065 


32 


1024 


32768 


5.6568542 


3.1748021 


.0312.50000 


33 


1089 


35937 


5.7445()2() 


3.2075^U3 


.0:30:30:50:30 


34 


1156 


39304 


5.8:509519 


3.2:306118 


.0294117(55 


35 


1225 


42875 


5.9160798 


3.2710663 


.028571429 


36 


1296 


46656 


6.0000000 


3.:5019272 


.027777778 


37 


1369 


50653 


6.0827(525 


3.:3:322218 


.027027(V27 


38 


1444 


54872 


6.1()44140 


3.:5(519754 


.026:515789 


39 


1521 


59319 


6.2449^)80 


3.3912114 


.025641026 


40 


1600 


64000 


6.3245553 


3.41<9519 


.02.")0noooo 


41 


1681 


68921 


6.4031242 


3.44S2172 


.024:3'. K)244 


42 


lliyi 


74088 


6.4807407 


3.47(i02(3(3 


.02:3S(MI524 


43 


1849 


79507 


6.5574:585 


3.50:5:5981 


.02:52.-)5S14 


44 


1936 


85184 


().():^:>249() 


3.530:5483 


.022727273 


45 


2025 


91125 


6.7()<S2():5<) 


3.55(),S0:33 


/)>)»)0*)'>')02 


4(5 


2116 


973:^ 


6.782:5:i()0 


3.58:30479 


!02173rtir3(") 


47 


2209 


103823 


().(S55().')4(> 


3.(50882(51 


.021 27t 5596 


48 


2304 


110592 


6.92S20:V2 


3.6:342411 


.0208:5:5:3:33 


49 


2401 


ii7(;4i) 


7.000()()00 


3.659:3057 


.O2O40Sl(53 



360 SQUARES, CUBES, SQUARE ROOTS, ETC. 

2 



SQUARES, CUBES 


, SQUARE ROOTS, 


CUBE 




EOOTS, AND RECIPROCALS 




Jfo. 


Squares 


Cubes 


Square Roots 


Cube Roots 


Reciprocals 


50 


2500 


125000 


7.0710678 


3.6840314 


.020000000 


51 


2601 


132651 


7.1414284 


3.7084298 


.019(J07843 


52 


2701 


140608 


7.2111026 


3.7325111 


.019230769 


53 


2809 


148877 


7.2801099 


3.7562858 


.018867925 


54 


2916 


157464 


7.3484692 


3.7797631 


.018518519 


55 


3025 


166375 


7.4161985 


3.8029525 


.018181818 


5(5 


3136 


175616 


7.48.33148 


3.8258624 


.017857143 


57 


3249 


185193 


7.5498344 


3.8485011 


.017543860 


58 


3364 


195112 


7.6157731 


3.870876() 


.017241379 


59 


3181 


205379 


7.6811457 


3.8929965 


.016949153 


60 


3600 


216000 


7.7459667 


3.9148676 


.016666667 


61 


3721 


226981 


7.8102497 


3.9364972 


.016393443 


62 


3814 


238328 


7.8740079 


3.9578915 


.016129032 


63 


3969 


250047 


7.9372539 


3.9790571 


.015873016 


64 


4096 


262144 


8.0000000 


4.0000000 


.015625000 


65 


4225 


274625 


8.0622577 


4.0207256 


.015384615 


66 


4356 


287496 


8.1240384 


4.0412401 


.015151515 


67 


4489 


300763 


8.1853528 


4.0615480 


.014925373 


68 


4624 


314432 


8.2462113 


4.0816551 


.014705882 


69 


4761 


328509 


8.3066239 


4.1015661 


.014492754 


70 


4900 


343000 


8.3G66003 


4.1212853 


.014285714 


71 


5041 


357911 


8.4261498 


4.1408178 


.014084507 


72 


5184 


373248 


8.4852814 


4.1601676 


.013888889 


73 


5329 


389017 


8.5440037 


4.1793390 


.013698630 


7i 


5476 


405224 


8.6023253 


4.1983364 


.013513514 


75 


5625 


421875 


8.6602540 


4.2171633 


.013333333 


76 


5776 


438976 


8.7177979 


4.2358236 


.013157895 


77 


5929 


456533 


8.7749644 


4.2543210 


.012987013 


78 


6084 


474552 


8.8317609 


4.2726586 


.012820513 


79 


6241 


493039 


8.8881944 


4.2908404 


.012658228 


80 


6400 


512000 


8.9442719 


4.3088695 


.012500000 


81 


6561 


531441 


9.0000000 


4.3267487 


.012345679 


82 


6724 


551368 


9.0553851 


43444815 


.012195122 


83 


6889 


571787 


9.1104336 


4.3620707 


.012048193 


84 


7056 


592704 


9.1651514 


4.3795191 


.0119047()2 


85 


7225 


614125 


9.2195445 


4.396821K) 


.011764706 


86 


7396 


636056 


9.2736185 


4.4140049 


.011(327907 


87 


7569 


658503 


9.3273791 


4.4310476 


.011494253 


88 


7744 


681472 


9.3808315 


4.4479602 


.011363636 


89 


7921 


7049()9 


9.4339811 


4.4647451 


.011235955 


90 


8100 


72^)000 


9.4868330 


4.4814047 


.011111111 


91 


8281 


753571 


9.5393920 


4.4979414 


.010989011 


92 


8464 


778688 


9.5f)16630 


4.5143574 


.010869565 


93 


8649 


804357 


9.6436508 


4.5.306549 


.010752688 


91 


8836 


8305S4 


9.695.3.^)97 


4.54()8359 


.010(538298 


95 


9025 


857;'.75 


9.74()7943 


4..")(;29026 


.01052(5316 


96 


9216 


88473() 


9.7979.'")90 


4.57S8.570 


.010416(5(57 


97 


9409 


912673 


9.848S578 


4.5947009 


.010309278 


98 


9()04 


941192 


9.8994949 


4.61043()3 


.010204082 


99 


9801 


970299 


9.9498744 


4.6260650 


.010101010 



8QUABES, CUBES, SQUARE SOOTS, ETC. 

3 



361 



SQUARES, CUBES 


, SQUARE ROOTS, 


CUBE 




ROOTS, AND RECIPROCALS 




No. 


Squares 


Cubes 


Square Roots 


Cube Roots 


Reriprorals 


100 


10000 


100(^000 


10.0000000 


4.(5415888 


.oiooooooo 


101 


10201 


1030301 


10.049875(5 


4.(5570095 


.009900990 


102 


10404 


1061208 


10.0995049 


4.^)72:3287 


.00i)8():i922 


103 


10G09 


1092727 


10.1488916 


4.(3875482 


.00970S7:}8 


104 


1081(3 


1124864 


10.1980390 


4.702(5(594 


.009(515:W5 


105 


11025 


1157(325 


10.24(59508 


4.717(5940 


.00952:5810 


10() 


11236 


1191016 


10.295(5301 


4.732(52:35 


.0094:5:59(52 


107 


11449 


1225043 


10.:5440804 


4.7474594 


.0(^)9:145794 


108 


1161)4 


1259712 


10.3923048 


4.7(5220:52 


.0(J9259259 


109 


11881 


1295029 


10.4403065 


4.7768562 


.009174312 


110 


12100 


1331000 


10.4880885 


4.7914199 


.009090909 


111 


12321 


13()7631 


10.535(5538 


4.8058955 


.009009009 


112 


12544 


1404928 


10.5830052 


4.8202845 


.008928571 


113 


127()9 


1442897 


10.6301458 


4.8:345881 


.008849558 


114 


1299() 


1481544 


10.6770783 


4.8488076 


.008771!):30 


115 


13225 


1520875 


10.7238053 


4.8(529442 


.008695(552 


IK) 


13456 


1560896 


10.7703296 


4.87(59990 


.008(520(590 


117 


13689 


1601(313 


10.8166538 


4.89097:32 


.008547009 


118 


13924 


1(343032 


10.8(527805 


4.9048(581 


.008474576 


119 


14161 


1685159 


10.9087121 


4.9186847 


.008403361 


120 


14400 


1728000 


10.9544512 


4.9324242 


.0083333:33 


121 


14(541 


1771561 


11.0000000 


4.94(50874 


.0082(544(53 


122 


14884 


1815848 


11.0453610 


4.9596757 


.00819(5721 


123 


15129 


18608()7 


11.09053(55 


4.9731898 


.0081:30081 


124 


15:'>7() 


1906624 


11.1355287 


4.98(5(5310 


.0080(5451(5 


125 


15()25 


1953125 


11.1803:599 


5.000(X)00 


.008000000 


12(i 


15876 


2000376 


11.2249722 


5.01:32979 


.0079:3(3508 


127 


16129 


2048383 


11.2(594277 


5.02(55257 


.00787401(5 


128 


16384 


2097152 


11.3137085 


5.0:39(5842 


.007812500 


121) 


1(3()41 


2146689 


11.3578167 


5.0527743 


.007751938 


130 


16900 


2197000 


11.4017543 


5.0657970 


.007692:308 


131 


171()1 


2248091 


11.4455231 


5.07875:51 


.0076:53588 


132 


17424 


22i)99()8 


11.4891253 


5.091(54:34 


.007575758 


133 


17689 


2352(i37 


11.5325(526 


5.1044(587 


.007518797 


134 


1795() 


240(5104 


11.5758369 


5.1172299 


.0074(52(587 


l.T) 


18225 


21(5():'>75 


11.6189500 


5.129H278 


.007407407 


i:;(j 


1849() 


2515456 


11.(5(5190:58 


5.1425(5:32 


.007:352941 


137 


18769 


2571353 


11.704(5999 


5.1551:3(57 


.007299270 


138 


11H)44 


2(52.S072 


11.747:3401 


5.1(57(5493 


.00724(5:577 


139 


19321 


2685(319 


11.7898261 


5.1801015 


.007194245 


140 


19(;00 


2744000 


11.8321596 


5.1924941 


.0071428.57 


141 


19881 


2803221 


11.874:3421 


5.2()4S279 


.007092 11 »9 


142 


20164 


28(5.".288 


ll.i)l(5.3753 


5.21710:34 


.0070422.54 


143 


20449 


2!)24207 


11.9582(507 


5.229:3215 


.00(599:5007 


144 


207;5() 


2985984 


12.(X)()00()0 


5.2414828 


.006944444 


145 


21025 


3048(525 


12.041594(5 


5.25:i5879 


.0O(589(55.")2 


146 


21316 


3112136 


12.0.S:;04(50 


5.2(55(5.374 


.00(5,S49315 


147 


21()()9 


317(5523 


12.124:5557 


5.277(5:321 


.()0(5SO2721 


148 


21904 


3241792 


12.1(55.-)251 


5.2895725 


.00(575(5757 


149 


22201 


's^mm 


12.20(5.")55() 


5.:5()14592 


.00(5711409 



362 SQUARES, CUBES, SQUARE ROOTS, ETC. 

4 



SQUARES, CUBES 


, SQUARE ROOTS, 


CUBE 




ROOTS, AND RECIPROCALS 




Ao. 


Squares 


Cul)('s 


Square Roots 


Cube Roots 


Reciprocals 


150 


22500 


3375000 


12.2474487 


5.3132928 


.0066()(5(i(57 


151 


22801 


3442951 


12.2882057 


5.3250740 


.00(5(522517 


152 


23104 


3511808 


12.3288280 


5.3368033 


.00657S947 


153 


23409 


3581577 


12.3693169 


5.3484812 


.006535948 


154 


23716 


3652264 


12.4096736 


5.3(301084 


.00(3493506 


155 


24025 


3723875 


12.4498996 


5.3716854 


.006451613 


156 


24336 


3796416 


12.4899960 


5.3832126 


.006410256 


157 


24649 


3869893 


12.5299641 


5.3946907 


.006369427 


158 


24964 


3944312 


12.5698051 


5.4061202 


.006329114 


159 


25281 


4019679 


12.6095202 


5.4175015 


.006289308 


ino 


25600 


4096000 


12.6491106 


5.4288352 


.006250000 


161 


25921 


4173281 


12.6885775 


5.4401218 


.006211180 


162 


26244 


4251528 


12.7279221 


5.4513618 


.006172840 


163 


26569 


4330747 


12.7671453 


5.4625556 


.0061349(59 


164 


26896 


4410944 


12.8062485 


5.4737037 


.006097561 


165 


27225 


4492125 


12.8452326 


5.4848066 


.00(5060(506 


166 


27556 


4574296 


12.8840987 


5.4958647 


.006024096 


167 


27889 


4(357463 


12.9228480 


5.5068784 


.005988024 


168 


28224 


4741632 


12.9614814 


5.5178484 


.005952381 


169 


28561 


4826809 


13.0000000 


5.5287748 


.005917160 


170 


28900 


4913000 


13.0384048 


5.5396583 


.005882353 


171 


29241 


5000211 


13.0766968 


5.5504991 


.005847953 


172 


29584 


5088448 


13.1148770 


5.5612978 


.005813953 


173 


29929 


5177717 


13.1529464 


5.5720546 


.005780347 


174 


30276 


5268024 


13.1909060 


5.5827702 


.005747126 


175 


30625 


5359375 


13.2287566 


5.5934447 


.005714286 


176 


30976 


5451776 


13.2(564992 


5.(3040787 


.005681818 


177 


31329 


5545233 


13.3041347 


5.6146724 


.005649718 


178 


31684 


5639752 


13.341(3(341 


5.6252263 


.005617978 


179 


32041 


5735339 


13.3790882 


5.6357408 


.005586592 


180 


32400 


5832000 


13.4164079 


5.(i4(321(32 


.005555556 


181 


32761 


5929741 


13.4536240 


5.6566528 


.0055248(52 


182 


33124 


6028568 


13.4907376 


5.6(570511 


.005494505 


183 


33489 


6128487 


13.5277493 


5.6774114 


.005464481 


184 


33856 


6229504 


13.564(3(300 


5.6877340 


.005434783 


185 


34225 


6331625 


13.(3014705 


5.6980192 


.005405405 


186 


34596 


6434856 


13.6381817 


5.7082675 


.00537(5344 


187 


34969 


6539203 


13.(3747943 


5.7184791 


.005347594 


188 


35344 


6644()72 


13.7113092 


5.728(5543 


.005319149 


189 


35721 


6751269 


13.7477271 


5.7387936 


.005291005 


190 


36100 


6859000 


13.7840488 


5-7488971 


.005263158 


191 


36481 


69()7871 


13.8202750 


5.7589652 


.005235602 


192 


3()864 


7077888 


13.85()40(35 


5.7(589982 


.005208333 


193 


37249 


7189057 


13.8924440 


5.77899(36 


.005181347 


194 


37636 


7301384 


13.9283883 


5.7889(304 


.005154639 


195 


38025 


7414875 


13.9642400 


5.7988900 


.005128205 


196 


38416 


7529536 


14.0000000 


5.8087857 


.005102041 


197 


38809 


7645373 


14.035(3(388 


5.818(3479 


.00507(5142 


198 


39204 


7762392 


14.0712473 


5.8284767 


.005050505 


199 


39601 


7880599 


14.10(373(30 


5.8382725 


.005025126 



SQUAIiES, CUBES, SQUAHE HOOTS, ETC. 



3G3 



SQUARES, CUBES 


, SQUARE ROOTS, 


CUBE 




ROOTS, AND RECIPROCALS 




No. 


Squares 


Cubes 


Square Roots 


Cube Roots 


Reciprorals 


200 


40000 


8000000 


14.1421356 


5.8480355 


.005(K)(K)(K) 


201 


40401 


8120601 


14.17744()9 


5.85776()0 


.004975121 


202 


40804 


8242408 


14.212()704 


5.8674643 


.0049.50195 


203 


41209 


8365427 


14.2478068 


5.8771307 


.004926108 


204 


41()16 


848VU;64 


14.28285()9 


5.8867653 


.00490 19()1 


205 


42025 


8(U5125 


14.3178211 


5.89()3685 


.004S7.S049 


206 


4243() 


8741816 


14.3527001 


5.90.-)9406 


.0048.54.369 


207 


42849 


88()9743 


14.3874946 


5.9154817 


.0048.30918 


208 


43264 


8998912 


14.4222051 


5.9249921 


.004807692 


209 


43681 


9129329 


14.4568323 


5.9344721 


.004784(389 


210 


44100 


9261000 


14.491.3767 


5.9439220 


.004761905 


211 


44521 


9393931 


14.525835)0 


5.9533418 


.0047:39:33f) 


212 


44944 


9528128 


14.5(i02198 


5.9627320 


.00471()981 


213 


453()9 


9663597 


14.5945195 


5.9720926 


.004()1K18:36 


214 


4579() 


9800344 


14.()287388 


5.9814240 


.004(572897 


215 


4()225 


9938375 


14.6628783 


5.99072(>4 


.004(5511(53 


216 


4()6r)6 


10077()96 


14.()969385 


6.0000000 


.004629(5:30 


217 


47089 


10218313 


14.7309199 


6.0092450 


.004(508295 


218 


47524 


10360232 


14.7648231 


6.0184()17 


.004.5871.56 


219 


471K51 


10503459 


14.7986486 


6.0276502 


.00456(5210 


220 


48400 


10648000 


14.8323970 


6.0368107 


.(X)4545455 


221 


48S41 


10793861 


14.8660687 


6.0459435 


.004524887 


222 


492.S4 


10941048 


14.8996644 


6.0550489 


.004.504505 


223 


49729 


11089567 


14.9331845 


6.0()41270 


.004484.305 


224 


50176 


11239424 


14.966()295 


6.0731779 


.0044(54286 


225 


50()25 


113<K)625 


15.0000000 


6.0822020 


.004444444 


226 


5107() 


11543176 


15.03: '>2964 


6.0911994 


.004424779 


227 


51529 


11697083 


15.06()5192 


6.1001702 


.004405286 


228 


51984 


11852352 


15.0996()89 


6.1091147 


.004:3859(55 


229 


52441 


12008989 


15.1327460 


6.1180332 


.004:3(56812 


230 


52900 


12167000 


15.1657509 


6.1269257 


.004.347826 


231 


53361 


1232()391 


15.1986842 


6.1357924 


.004329004 


232 


53824 


12487168 


15.2315462 


6.14463.37 


.004310:U5 


233 


54289 


12649337 


15.2(543.375 


6.1.5.34495 


.004291845 


2.34 


54756 


12812904 


15.2970585 


6.1622401 


.00427:3.504 


235 


55225 


12977875 


15.32970i)7 


6.1710058 


.004255319 


236 


55696 


13144256 


15.3622915 


6.1797466 


.0042:>72S8 


237 


56169 


13312053 


15.3948043 


6.1884628 


.004219409 


238 


56644 


13481272 


15.4272486 


6.1971544 


.004201(581 


239 


57121 


13()51919 


15.4r)96248 


6.20.-).S218 


.004 1841 (Ml 


240 


57600 


13824000 


15.4919.3.34 


6.2144650 


.004 1(3(5(5(37 


241 


58081 


13997521 


15.5241747 


6.22:30843 


.004149:^78 


242 


58564 


14172488 


15.5."56:U92 


6.2316797 


.0041:^2231 


243 


51K)49 


14:H8907 


15.5884573 


6.2402515 


.00411.5226 


244 


59536 


14526784 


15.()204994 


6.2487998 


.004098:3(51 


245 


60025 


1470(;i25 


15.6.")24758 


6.2.')7:'»248 


.004081(5:^3 


246 


6051() 


14<SS69:36 


15.6S4."»S71 


6.2().~)82()6 


.0040(5.5041 


247 


61009 


15069223 


15.71 62.3:U) 


(). 274:3054 


.00404S583 


248 


61504 


15252992 


15.74S01.57 


(). 28276 13 


.0040:52258 


249 


62001 


15438249 


15.7797338 


6.2911946 


.00401(5(^54 



364 SQUARES, CUBES, SQUARE ROOTS, ETC. 

6 



SQUARES, CUBES 


, SQUARE ROOTS, 


CUBE 




ROOTS, AND RECIPROCALS 




Xo. 


Squares 


Cubes 


Square Roots 


Cube Roots 


Reciprocals 


250 


62500 


15625000 


15.8113883 


6.2996053 


.004000000 


251 


63001 


15813251 


15.8429795 


6.3079935 


.003984064 


252 


63504 


16003008 


15.8745079 


6.3163596 


.0039(58254 


253 


64009 


16194277 


15.9059737 


6.3247035 


.003952569 


254 


64516 


16387064 


15.9373775 


6.3330256 


.003937008 


255 


65025 


16581375 


15.9687194 


6.3413257 


.003921569 


256 


65536 


16777216 


16.0000000 


6.3496042 


.003906250 


257 


66049 


16974593 


16.0312195 


6.3578611 


.003891051 


258 


66564 


17173512 


16.0623784 


6.3660968 


.003875969 


259 


67081 


17373979 


16.0934769 


6.3743111 


.003861004 


260 


67600 


17576000 


16.1245155 


6.3825043 


.003846154 


261 


68121 


17779581 


16.1554944 


6.3906765 


.003831418 


262 


68644 


17984728 


16.1864141 


6.3988279 


.003816794 


263 


691()9 


18191447 


16.2172747 


6.4069585 


.003802281 


264 


69696 


18399744 


16.2480768 


6.4150687 


.003787879 


265 


70225 


18()09625 


16.2788206 


6.4231583 


.003773585 


266 


70756 


18821096 


16.3095064 


6.4312276 


.003759398 


267 


71289 


19034163 


16.3401346 


6.4392767 


.003745318 


268 


71824 


19248832 


16.3707055 


6.4473057 


.003731343 


269 


723(U 


19465109 


16.4012195 


6.4553148 


.003717472 


270 


72900 


19683000 


16.4316767 


6.4633041 


.003703704 


271 


73441 


19902511 


16.4620776 


6.4712736 


.003690037 


272 


73984 


20123<)48 


16.4924225 


6.4792236 


.003676471 


273 


74529 


20346417 


16.5227116 


6.4871541 


.003663004 


274 


75076 


20570824 


16.5529454 


6.4950653 


.003649635 


275 


75625 


20796875 


16.5831240 


6.5029572 


.003636364 


276 


76176 


21024576 


16.6132477 


6.5108300 


.003623188 


277 


76729 


21253933 


16.6433170 


6.5186839 


.003610108 


278 


77284 


21484952 


16.6733320 


6.5265189 


.003597122 


279 


77841 


21717639 


16.7032931 


6.5343351 


.003584229 


280 


78400 


21952000 


16.7332005 


6.5421326 


.003571429 


281 


78961 


22188011 


16.7630546 


6.5499116 


.003558719 


282 


79524 


22425768 


16.7928556 


6.5576722 


.00354()099 


283 


80089 


22665187 


16.822()038 


6.5654144 


.003533569 


284 


80656 


22906304 


16.8522995 


6.5731385 


.003521127 


285 


81225 


23149125 


16.8819430 


6.5808443 


.003508772 


286 


81796 


23393656 


16.9115345 


6.5885323 


.003496503 


287 


82369 


23639903 


16.9410743 


6.5962023 


.003484321 


288 


82944 


23887872 


16.9705627 


6.6038545 


.003472222 


289 


83521 


24137569 


17.0000000 


6.6114890 


.003160208 


290 


84100 


24389000 


17.0293864 


6.6191060 


.003448276 


291 


84681 


24642171 


17.0587221 


6.62()7054 


.00343(5426 


292 


85264 


24897088 


17.0880075 


().6342874 


.003424(558 


293 


85849 


25153757 


17.1172428 


6.6418522 


.003412969 


294 


86436 


25412184 


17.1464282 


6.6493998 


.003401361 


295 


87025 


25672375 


17.1755640 


6.(i5()9302 


003389831 


296 


87616 


25934336 


17.204(5505 


6.6644437 


.003378378 


297 


88209 


26198073 


17.23:;6S79 


().6719403 


.0033(57003 


298 


88804 


2CAiyM)2 


17.2626765 


6.(5794200 


.003355705 


299 


89401 


267:50899 


17.2916165 


6.6868831 


.003344482 



SQUARES, CUBES, SQUARE ROOTS, ETC. 365 

7 



SQUARES, CUBES 


, SQUARE ROOTS, 


CUBE 




ROOTS, AND RECIPROCALS 




No. 


Squares 


Cubes 


Square Roots 


Cube Roots 


Reriprorals 


300 


90000 


27000000 


17.3205081 


6.(5943295 


.003:5:^3:533 


301 


90601 


27270901 


17.349351() 


6.7017593 


.003322259 


302 


91204 


27543608 


17.3781472 


6.7091729 


.003311258 


303 


91809 


27818127 


17.40()8952 


6.71(55700 


.003300:5:50 


30i 


9241() 


280944()4 


17.4355958 


6.72:39508 


.00:5289474 


305 


93025 


2S372()25 


17.4(542492 


6.7313155 


.00:5278(581 


306 


936;^>() 


28()52r)16 


17.4928557 


6.738(5(541 


.0032(57974 


307 


94249 


28934443 


17.5214155 


6.7459967 


.003257329 


308 


948(54 


29218112 


17.5499288 


6.7533134 


.00324675:5 


309 


95481 


29503629 


17.5783958 


6.7606143 


.00323624(5 


310 


96100 


29791000 


17.(50(58169 


6.7(578995 


.00322580(5 


311 


9()721 


300802;U 


17.6351921 


6.7751(590 


.0032154:^ 


312 


97344 


30371328 


17.6635217 


6.7824229 


.003205128 


313 


97969 


30664297 


17.69180(50 


6.789(5613 


.003194888 


314 


98596 


30959144 


17.7200451 


6.7968844 


.003184713 


315 


99225 


31255875 


17.7482393 


(5.8040921 


.003174(503 


316 


9985() 


31554496 


17.77(53888 


6.8112847 


.003164557 


317 


100489 


31855013 


17.8044938 


6.8184620 


.003154574 


318 


101124 


32157432 


17.8325545 


6.825(5242 


.003144r,.54 


319 


101761 


32461759 


17.8605711 


6.8327714 


.003134796 


320 


102400 


32768000 


17.8885438 


6.8399037 


.003125000 


321 


103041 


33076161 


17.9164729 


6.8470213 


.0031152(55 


322 


103684 


33386248 


17.9443584 


6.8541240 


.0031055^)0 


323 


104329 


33698267 


17.9722008 


6.8612120 


.003095975 


324 


104976 


34012224 


18.0000000 


6.8682855 


.00308(5420 


325 


105625 


34328125 


18.0277564 


6.875:5443 


.00:307(5923 


326 


10(L>76 


34645976 


18.0554701 


6.8823888 


.00:50(57485 


327 


106929 


349(35783 


18.0831413 


6.8894188 


.00:5058104 


328 


107584 


35287552 


18.1107703 


6.89(54:545 


.003048780 


329 


108241 


35611289 


18.1383571 


6.9034359 


.003039514 


330 


108900 


35937000 


18.1659021 


6.9104232 


.0030:50:503 


331 


1095()1 


362(54691 


18.1934054 


6.917:3964 


.00:3021148 


332 


110224 


3()594;^)8 


18.2208(572 


6.9243556 


.003012048 


333 


110889 


36926037 


18.248287(5 


6.931:5008 


.00:500:500:5 


334 


111556 


37259704 


18.275(5(5(59 


6.9:582:321 


.002994012 


3:35 


112225 


37595375 


18.3030052 


6.9451496 


.002985075 


336 


112896 


37933056 


18.330:5028 


6.9520533 


.00297(51<X) 


337 


113569 


38272753 


18.3575598 


6.95894:54 


.0029(57:559 


338 


114244 


38614472 


18.3847763 


6.^X558198 


.002958580 


339 


114921 


38958219 


18.4119526 


6.9726826 


.002^985:5 


340 


115600 


39304000 


18.4390889 


6.9795.321 


.002941176 


341 


116281 


39651821 


18.4(561853 


6.98(53681 


.0029:52551 


342 


1169()4 


40001688 


18.4932420 


6.9931906 


.00292:5977 


343 


117()49 


40;i5;5()07 


18.5202592 


7.0(X)0000 


.002915452 


344 


118336 


40707584 


18.5472370 


7.00(579(52 


.0021K)(5977 


345 


119025 


^lOiVMVIo 


18.5741756 


7.01:35791 


.002898551 


346 


119716 


41421736 


18.(5010752 


7.02(\34<)0 


.0O28<)0173 


347 


120409 


417S1923 


18.(5279:1(50 


7.0271058 


.002,S81844 


348 


121104 


42144192 


18.(5.">47581 


7.0:5:58497 


.00287:5.-)(53 


349 


121801 


42508549 


18.(5815417 


7.0405806 


.0028653:30 



366 SQUABE8, CUBES, SQUARE ROOTS, ETC. 

8 



SQUARES, CUBES 


, SQUARE ROOTS, 


CUBE 




ROOTS, AND RECIPROCALS 




No. 


Squares 


Cubes 


Square Roots 


Cube Roots 


Reciprocals 


350 


122500 


42875000 


18.7082869 


7.0472987 


.002857143 


351 


123201 


43243551 


18.734V)910 


7.0540041 


.002849003 


352 


123904 


43614208 


18.761(5630 


7.0006967 


.002840909 


353 


124609 


43986977 


18.7882942 


7.0673767 


.002832861 


354 


125316 


443()1864 


18.8148877 


7.0740440 


.002824859 


355 


126025 


44738875 


18.8414437 


7.0806988 


.002816901 


356 


126736 


45118016 


18.8679623 


7.0873411 


.002808989 


357 


127449 


45499293 


18.8944436 


7.0939709 


.002801120 


358 


128164 


45882712 


18.92088V9 


7.1005885 


.002793296 


359 


128881 


46268279 


18.9472953 


7.1071937 


.002785515 


360 


129600 


46656000 


18.9736660 


7.1137866 


.002777778 


361 


130321 


47045881 


19.0000000 


7.1203674 


.002770083 


362 


131044 


47437928 


19.0262976 


7.1269360 


.002762431 


363 


131769 


47832147 


19.0525589 


7.1334925 


.002754821 


364 


132496 


48228544 


19.0787840 


7.1400370 


.002747253 


365 


133225 


48627125 


19.1049732 


7.1465695 


.002739726 


366 


133956 


49027896 


19.1311265 


7.1530901 


.002732240 


367 


134689 


49430863 


19.1572441 


7.1595988 


.002724796 


368 


135424 


49836032 


19.1833261 


7.1660957 


.002717391 


369 


136161 


50243409 


19.2093727 


7.1725809 


.002710027 


370 


136900 


50653000 


19.2353841 


7.1790544 


.002702703 


371 


137641 


51064811 


19.2613603 


7.1855162 


.002695418 


372 


138384 


51478848 


19.2873015 


7.1919663 


.002688172 


373 


139129 


51895117 


19.3132079 


7.1984050 


.002680965 


374 


139876 


52313624 


19.3390796 


7.2048322 


.002673797 


375 


140625 


52734375 


19.3649167 


7.2112479 


.002666667 


376 


141376 


53157376 


19.3907194 


7.2176522 


.002659574 


377 


142129 


53582633 


19.4164878 


7.2240450 


.002652520 


378 


142884 


54010152 


19.4422221 


7.2304268 


.0026)45503 


379 


143641 


54439939 


19.4679223 


7.2367972 


.002638522 


380 


144400 


54872000 


19.4935887 


7.2431565 


.002631579 


381 


145161 


55306341 


19.5192213 


7.2495045 


.002624672 


382 


145924 


55742968 


19.5448203 


7.2558415 


.002617801 


383 


146689 


56181887 


19.5703858 


7.2621675 


.002()10966 


384 


147456 


56623104 


19.5959179 


7.2684824 


.002604167 


385 


148225 


57066625 


19.6214169 


7.2747864 


.002597403 


386 


148996 


57512456 


19.6468827 


7.2810794 


.002590674 


387 


149769 


57960603 


19.6723156 


7.2873()17 


.002583979 


388 


150544 


58411072 


19.6977156 


7.2936330 


.002577320 


389 


151321 


58863869 


19.7230829 


7.2998936 


.002570694 


390 


152100 


59319000 


19.7484177 


7.3061436 


.002564103 


391 


152881 


59776471 


19.7737199 


7.3123828 


.002557545 


392 


153(J64 


60236288 


19.7989S99 


7.3186114 


.002551020 


393 


154149 


60()98457 


19.8242276 


7.3248295 


.002544529 


394 


155236 


611()2984 


19.8494332 


7.33103()9 


.002538071 


395 


156025 


61()29875 


19.874()069 


7.3372339 


.002531646 


396 


156816 


62099136 


19.8997487 


7.3434205 


.002525253 


397 


157609 


62570773 


19.9248588 


7.3495966 


.002518892 


398 


158404 


63044792 


19.9499373 


7.35.57()24 


.002512563 


399 


159201 


63521199 


19.9749844 


7.3619178 


.0025062()() 



SQUARES, CUBES, SQUARE ROOTS, ETC. 

9 



367 



SQUARES, CUBES 


, SQUARE ROOTS, 


CUBE 




ROOTS, AND RECIPROCALS 




Itfo. 


Squares 


Cabes 


Square Roots 


Cube Roots 


Reriprorals 


400 


160000 


(;40ouooo 


20.0000000 


7.3680()30 


.0()2r)()()0()o 


401 


1()0S01 


64481201 


20.0249844 


7.3741979 


.00249:576(5 


402 


161604 


649(>4808 


20.0499377 


7.3803227 


.0024875(52 


403 


162409 


65450827 


20.0748599 


7.3864373 


.002481:590 


404 


163216 


65939264 


20.0997512 


7.3925418 


.002475248 


405 


1()4025 


664:^125 


20.1246118 


7.398()362 


.0024(591:5(5 


406 


16483() 


66923416 


20.1494417 


7.4047206 


.0024(5:5054 


407 


165()49 


67419143 


20.1742410 


7.4107950 


.002457002 


408 


16()464 


67917312 


20.19rK)099 


7.4168595 


.0024.-)0980 


409 


167281 


68417929 


20.2237484 


7.4229142 


.002444988 


410 


168100 


68921000 


20.24S45()7 


7.4289589 


.0024:59024 


411 


168921 


69426531 


20.2731349 


7.4349938 


.0024:5:5090 


412 


169744 


69934528 


20.2977831 


7.4410189 


.002427184 


413 


170569 


70444997 


20.3224014 


7.4470:542 


.002421:508 


414 


171396 


70957944 


20.34()9899 


7.45:>0:399 


.002415459 


415 


172225 


71473375 


20.3715488 


7.4590359 


.002409(5:59 


416 


17305() 


71991296 


20.3900781 


7.4650223 


.00240:5846 


417 


173889 


72511713 


20.4205779 


7.4709991 


.002398082 


418 


174724 


73034632 


20.4450483 


7.4769(564 


.002392:344 


419 


175561 


73560059 


20.4694895 


7.4829242 


.002:5866:55 


420 


176400 


74088000 


20.4939015 


7.4888724 


.002380952 


421 


177241 


74618461 


20.5182845 


7.494S113 


.002375297 


422 


178084 


75151448 


20.542()386 


7.5007406 


.002:5(59(5(58 


423 


178929 


7568()iK57 


20.5669()38 


7.50(5(5(507 


.002:5640(56 


424 


179776 


76225024 


20.5912603 


7.5125715 


.002:558491 


4'^ 


180625 


767()5625 


20.()155281 


7.51847:30 


.002:552941 


426 


181476 


77308776 


20.6397()74 


7.524:3(552 


.002:547418 


427 


182329 


77854483 


20.6639783 


7.5:302482 


.002341920 


428 


183184 


78402752 


20.()881609 


7.5:561221 


.0023:5(5449 


429 


184041 


78953589 


20.7123152 


7.5419867 


.002331002 


430 


184900 


79507000 


20.7364414 


7.5478423 


.002:525581 


431 


185761 


80062991 


20.7605395 


7.55:36888 


.002:520186 


432 


186624 


806215()8 


20.7846097 


7.55952(53 


.002314815 


4:53 


187489 


81182737 


20.8()8()520 


7.565:3548 


.002:5(n>4(59 


434 


18835() 


81746504 


20.8326(;()7 


7.5711743 


.002:5(^4147 


435 


189225 


82312875 


20.85665:36 


7.57(59849 


.002298851 


43() 


1900i)() 


82881856 


20.880(5130 


7.58278(55 


.00229:5578 


437 


1<M)9(;9 


8:U53453 


20.<K)45450 


7.5885793 


.002288:5:50 


438 


191844 


84027672 


20.9284495 


7.5943(533 


.002283105 


439 


192721 


84604519 


20.952326)8 


7.(5001385 


.002277W4 


440 


193600 


85184000 


20.97()1770 


7.605<X)49 


.002272727 


441 


194481 


8576()121 


21.0000000 


7.611(5(52(5 


.0022(57574 


442 


1953()4 


8f)350888 


21.0237960 


7.617411(5 


.0022(5244:5 


443 


196249 


86938307 


21.0475652 


7.62:51519 


.002257:5:56 


444 


197136 


87528384 


21.07 1;'>075 


7.(52888:57 


.002252252 


445 


198025 


88121125 


21.()9.~)()231 


7.6:54(50(57 


.002247191 


446 


198<)1() 


8871()536 


21.1187121 


7.(540:3213 


.002242 1.")2 


447 


199809 


89314()23 


21.1423745 


7.(^(50272 


.0022:571:5(5 


448 


200704 


89915392 


21.1660105 


7.6517247 


.0022:52143 


449 


201601 


90518849 


21.1896201 


7.(55741:58 


.002227171 



368 SQUARES, CUBES, SQUARE BOOTS, ETC. 

10 



SQUARES, CUBES 


, SQUARE ROOTS, 


CUBE 




ROOTS, AND RECIPROCALS 




No. 


Squares 


Cubes 


Square Roots 


Cube Roots 


Reciprocals 


450 


202500 


91125000 


21.2132034 


7.6630943 


.002222222 


451 


20;i401 


91733851 


21.23()7606 


7.()687()()5 


;002217295 


452 


204304 


92345408 


21.2602916 


7.6744303 


.002212389 


453 


205209 


92959677 


21.28379(;7 


7.()800857 


.002207506 


454 


206116 


93576664 


21.3072758 


7.6857328 


.002202643 


455 


207025 


94196375 


21.3307290 


7.691.3717 


.002197802 


456 


207936 


94818816 


21.35415()5 


7.()970023 


.002192982 


457 


208849 


95443993 


21.3775583 


7.7026246 


.002188184 


458 


209764 


96071912 


21.4009346 


7.7082388 


.002183406 


459 


210681 


96702579 


21.4242853 


7.7138448 


.002178649 


460 


211600 


97336000 


21.4476106 


7.7194426 


.002173913 


461 


212521 


97972181 


21.4709106 


7.7250325 


.002169197 


462 


213444 


98611128 


21.4941853 


7.7306141 


.0021(14502 


463 


214369 


99252847 


21.5174348 


7.7361877 


.002159827 


464 


215296 


99897344 


21.5406592 


7.7417532 


.002155172 


465 


216225 


100544625 


21.5638587 


7.7473109 


.0021.50538 


466 


217156 


101194696 


21.5870331 


7.7528606 


.002145923 


467 


218089 


101847563 


21.6101828 


7.7584023 


.002141328 


468 


219024 


102503232 


21.6333077 


7.7639361 


.002i;5()752 


469 


219961 


103161709 


21.6564078 


7.7694620 


.002132196 


470 


220900 


103823000 


21.6794834 


7.7749801 


.002127660 


471 


221841 


104487111 


21.7025344 


7.7804904 


.002123142 


472 


222784 


105154048 


21.7255610 


7.7859928 


.002118644 


473 


223729 


105823817 


21.7485632 


7.7914875 


.002114165 


474 


224676 


106496424 


21.7715411 


7.7969745 


.002109705 


475 


225(525 


107171875 


21.7944947 


7.8024538 


.0021()52()3 


476 


226576 


107850176 


21.8174242 


7.8079254 


.002100840 


477 


227529 


108531333 


21.8403297 


7.8133892 


.0020!i()43() 


478 


228484 


109215352 


21.8632111 


7.8188456 


.002092050 


479 


229441 


109902239 


21.8860686 


7.8242942 


.002087683 


480 


230400 


110592000 


21.9089023 


7.8297353 


.002083333 


481 


231361 


in284()41 


21.9317122 


7.8351688 


.002079002 


482 


232324 


111980168 


21.9544984 


7.8405949 


.002074689 


483 


233289 


112678587 


21.9772610 


7.8460134 


.002070393 


484 


2:^256 


113379904 


22.0000000 


7.8514244 


.0020(5(5116 


485 


235225 


114084125 


22.0227155 


7.8.5{)8281 


.002061856 


486 


236196 


114791256 


22.0454077 


7.8()22242 


.002057(513 


487 


237169 


115501303 


22.0680765 


7.8676130 


.002053388 


488 


238144 


116214272 


22.0^)07220 


7.8729944 


.002049180 


489 


239121 


116930169 


22.1133444 


7.8783684 


.002044990 


490 


240100 


117649000 


22.13594.36 


7.8837352 


.002040816 


491 


241081 


118370771 


22.1585198 


7.8890946 


.002036(5(50 


492 


2420()4 


119095488 


22.1810730 


7.8944468 


.002032520 


4<)3 


24:^)049 


119823157 


22.2036033 


7.8997917 


.002028398 


494 


24403() 


120553784 


22.2261108 


7.9051294 


.002024291 


495 


245025 


121287375 


22.248.5955 


7.9104599 


.002020202 


496 


246016 


122023936 


22.2710575 


7.9157832 


.002016129 


497 


247009 


12276;M73 


22.29IU9()8 


7.92109()4 


.002012072 


498 


248004 


123505992 


22.31591 :U) 


7.9264085 


.0020080.32 


499 


249001 


124251499 


22.338:3079 


7.9317104 


.002004008 



SQUARES, CUBES, SQUARE ROOTS, ETC. 3G9 

11 



SQUARES, CUBES 


, SQUARE ROOTS, 


CUBE 




ROOTS, AND RECIPROCALS 




5o. 


Squares 


Cubes 


Square Roots 


Cube Roots 


Reciprocals 


500 


250000 


125000000 


22.360(;798 


7.9:370053 


.002000000 


501 


251001 


125751501 


22.3830293 


7.9422!>31 


.00199(5008 


502 


252004 


12()5()()008 


22.4053565 


7.94757:39 


.0019920:52 


503 


25:3009 


127263527 


22.427()615 


7.9528477 


.001988072 


504 


2:401() 


128024061 


22.4499443 


7.9581144 


.001984127 


505 


255025 


128787625 


22.4722051 


7.9(5:3:3743 


.001980198 


50() 


25()036 


12i)554216 


22.4944438 


7.9(58(5271 


.00197(5285 


507 


257049 


130323843 


22.51()6605 


7.97:38731 


.001972.387 


508 


2580()4 


131096512 


22.5388553 


7.9791122 


.0019(585(4 


509 


251K)81 


131872229 


22.5610283 


7.984:3444 


.0019(346:37 


510 


260100 


132()51000 


22.5831796 


7.9895(597 


.0019(50784 


511 


261121 


133432831 


22.(3053091 


7.9917883 


.00195(3147 


512 


262144 


134217728 


22.(3274170 


8.0000000 


.001953125 


513 


2()3169 


135005697 


22.(495033 


8.0052049 


.001949318 


514 


264196 


1357i)6744 


22.(3715(381 


8.01(40:32 


.001945525 


515 


265225 


13(3590875 


22.(5936114 


8.015594(5 


.001941748 


51(3 


266256 


137388096 


22.71563134 


8.0207794 


.0019:37984 


517 


2(57289 


138188413 


22.737(3340 


8.0259574 


.0019342:36 


518 


268324 


138991832 


22.75fK3134 


8.0:311287 


.0019.30502 


519 


269361 


139798359 


22.7815715 


8.0:362935 


.001926782 


520 


270400 


140608000 


22.8035085 


8.0414515 


.00192:3077 


521 


271441 


141420761 


22.8254244 


8.04(5(30:30 


.001919386 


522 


272484 


14223()648 


22.8473193 


8.0517479 


.00191.5709 


523 


273529 


143055(367 


22.8691933 


8.0568862 


.001912046 


524 


274576 


143877824 


22.8910463 


8.0(520180 


.0O11KK397 


525 


275()25 


144703125 


22.9128785 


8.0(371432 


.0019047(52 


52() 


27(>()76 


145531576 


22.934(5899 


8.0722(520 


.0011K)1141 


527 


277729 


14(33(33183 


22.9564806 


8.0773743 


.0018975:33 


528 


278784 


147197952 


22.9782506 


8.0824800 


.00189:3939 


529 


279841 


148035889 


23.0000000 


8.0875794 


.001890:359 


530 


280! )00 


148877000 


23.0217289 


8.092(5723 


.00188(5792 


531 


281961 


149721291 


23.04:4372 


8.0<)77589 


.00188:32.39 


532 


283024 


1505687(J8 


23.0(351252 


8. 1028; 590 


.001879(599 


533 


281089 


151419437 


23.0867928 


8.1079128 


.00187(5173 


5;u 


2851 5() 


152273304 


23.1084400 


8.112i)S03 


.001872(559 


535 


286225 


153130375 


23.1:300(570 


8.1180414 


.0018(59159 


536 


28729() 


1539<)0656 


23.151(3738 


8.12:309(32 


.0(M 8(55(572 


537 


2883()9 


154854153 


23.17:52(305 


8.1281447 


.0018(52197 


538 


289444 


155720872 


23.1948270 


8. 1:3:51 870 


.0018587:3(3 


539 


290521 


156590819 


23.21(53735 


8.1:3822:30 


.001855288 


540 


291(300 


157464000 


23.2:579001 


8.1432529 


.001851852 


541 


292681 


158340421 


23.25140(37 


8.14827(55 


.001.S4S429 


542 


2937()4 


159220088 


23.28089:35 


8.15:329:39 


.00184.1018 


543 


294849 


16010; ;oo7 


23.:302:5(501 


8.158:3051 


.001841(521 


544 


2!>59;'.6 


1()0989184 


23.:32:38076 


8.16:33102 


.(^) 18:382.35 


545 


2971)25 


I(;i878()25 


23.:452:\51 


8.1(58:3092 


.0018:48(52 


546 


298116 


162771336 


23.:3( 5(3(5429 


8.17:3;5020 


.fH)1831.n02 


547 


299209 


1().3()67323 


23.. 38803 11 


8.17S28S8 


.001S28J.14 


.548 


300304 


1(4566592 


23.409:3998 


8.18:32(595 


.001824818 


-49 


301401 


1()54(J<)149 


23.4:30741K) 


8.1882441 


.001821494 



2 b 



870 SQUABES, CUBES, SQUARE BOOTS, ETC. 

12 



SQUARES, CUBES 


, SQUARE ROOTS, 


CUBE 




ROOTS, AND RECIPROCALS 




No. 


Squares 


Cubes 
166375000 


Square Roots 


Cube Roots 


Reciprocals 


550 


302500 


23.4520788 


8.1932127 


.001818182 


551 


303601 


167284151 


23.4733892 


8.1981753 


.001814882 


552 


304704 


168196608 


23.4946802 


8.2031319 


.001811594 


553 


305809 


169112377 


23.5159520 


8.2080825 


.001808318 


554 


306916 


170031464 


23.5372046 


8.2130271 


.001805054 


555 


308025 


170953875 


23.5584380 


8.2179(357 


.001801802 


556 


309136 


171879616 


23.5796522 


8.2228985 


.001798561 


557 


310249 


172808693 


23.6008474 


8.2278254 


.001795332 


558 


311364 


173741112 


23.6220236 


8.2327463 


.001792115 


559 


312481 


174676879 


23.6431808 


8.2376614 


.001788909 


560 


313600 


175616000 


23.6643191 


8.2425706 


.001785714 


561 


314721 


176558481 


23.6854386 


8.2474740 


.061782531 


5()2 


315844 


177504328 


23.7065392 


8.2523715 


.001779359 


563 


316969 


178453547 


23.7276210 


8.2572633 


.001776199 


564 


318096 


179406144 


23.7486842 


8.2621492 


.001773050 


565 


319225 


180362125 


23.7697286 


8.2670294 


.001769912 


566 


320356 


181321496 


23.7907545 


8.2719039 


.00176()784 


567 


321489 


182284263 


23.8117618 


8.2767726 


.001763668 


568 


322624 


183250432 


23.8327506 


8.2816355 


.001760563 


569 


323761 


184220009 


23.8537209 


8.2864928 


.0017574(^9 


570 


324900 


185193000 


23.8746728 


8.2913444 


.001754386 


571 


326041 


186169411 


23.895(^063 


8.296^)03 


.001751313 


572 


327184 


187149248 


23.91(55215 


8.3010304 


.001748252 


573 


328329 


188132517 


23.9374184 


8.3058651 


.001745201 


574 


329476 


189119224 


23.9582971 


8.3106941 


.001742160 


575 


330625 


190109375 


23.9791576 


8.3155175 


.001739130 


576 


331776 


191102976 


24.0000000 


8.3203353 


.001736111 


577 


332929 


192100033 


24.0208243 


8.3251475 


.001733102 


578 


334084 


193100552 


24.0416306 


8.3299542 


.001730104 


579 


335241 


194104539 


24.0624188 


8.3347553 


.001727116 


580 


336400 


195112000 


24.0831891 


8.3395509 


.001724138 


581 


337561 


196122941 


24.1039416 


8.3443410 


.001721170 


582 


338724 


197137368 


24.12467(^2 


8.3491256 


.001718213 


583 


339889 


198155287 


24.1453929 


8.3539047 


.001715266 


584 


^41056 


199176704 


24.1(^(50919 


8.358(5784 


.001712329 


585 


342225 


200201625 


24.1867732 


8.3634466 


.001709402 


586 


M339f) 


201230056 


24.2074369 


8.3682095 


.001706485 


587 


344569 


202262003 


24.2280829 


8.3729(568 


.001703578 


588 


345744 


203297472 


24.2487113 


8.3777188 


.001700680 


589 


346921 


204336469 


24.2693222 


8.3824(553 


.001697793 


590 


348100 


205379000 


24.2899156 


8.38720(55 


.001(594915 


591 


349281 


206425071 


24.3104916 


8.3919423 


.001692047 


592 


350464 


207474688 


24.3310501 


8.39(5(5729 


.001689189 


593 


351649 


208527S57 


24.351.5913 


8.4013981 


.001686341 


594 


352836 


209584584 


24.3721152 


8.4061180 


.001683502 


595 


354025 


210(544875 


24.3926218 


8.410832(5 


.001(580672 


596 


355216 


211708736 


24.4131112 


8.4155419 


.001677852 


597 


356409 


21277(5173 


24.4335834 


8.42024(50 


.001675042 


598 


357604 


213847192 


24.4540385 


8.4249448 


.001()72241 


599 


358801 


214921799 


24.4744705 


8.429(5383 


.001(5(59449 



SQUARES, CUBES, SQUARE ROOTS, ETC. 

13 



371 



SQUARES, CUBES 


SQUARE ROOTS, 


CUBE 




ROOTS, AND RECIPROCALS 




No. 


Squares 


Cubes 


Square Roots 


Cube Roots 


Reciprocals 


600 


360000 


216000000 


24.4948974 


8.4343267 


.001(5(5(3(3(57 


601 


361201 


217081801 


24.5153013 


8.4390098 


.001(5(5:3894 


602 


362404 


218167208 


24.53515883 


8.443(5877 


.(X)l (5(51 1.30 


603 


363609 


219256227 


24.55(50583 


8.4483(305 


.001(558375 


604 


364816 


220348864 


24.5764115 


8.4530281 


.001655(529 


605 


36(5025 


221445125 


24.5967478 


8.457(5906 


.001(552893 


606 


367236 


222545016 


24.6170(373 


8.462:3479 


.001(5501(35 


607 


3()8449 


223648543 


24.6373700 


8.4670001 


.001(54744(5 


608 


36^)664 


224755712 


24.(i576560 


8.4716471 


.001644737 


609 


370881 


225866529 


24.6779254 


8.4762892 


.001(542036 


610 


372100 


226981000 


24.()981781 


8.4809261 


.001639:344 


611 


373321 


228099131 


24.7184142 


8.4855579 


.001(3:3(3(5(51 


612 


374544 


229220928 


24.738(3338 


8.4f)01848 


.001(333987 


613 


375769 


230346397 


24.7588368 


8.4948065 


.001631321 


614 


376996 


231475544 


24.7790234 


8.4994233 


.001628(3(54 


615 


378225 


232608375 


24.7991935 


8.5040350 


.001(52(3016 


616 


379456 


233744896 


24.8193473 


8.508(5417 


.001623377 


617 


380689 


234885113 


24.8394847 


8.5132435 


.001620746 


618 


381924 


23()029032 


24.8596058 


8.5178403 


.001618123 


619 


383161 


237176659 


24.8797106 


8.5224321 


.001615509 


620 


384400 


238328000 


24.8997992 


8.5270189 


.001612903 


621 


385641 


239483061 


24.9198716 


8.531(3009 


.001610:306 


622 


386884 


240641848 


24.9399278 


8.5361780 


.001(307717 


623 


388129 


241804367 


24.9599679 


8.5407501 


.001(5051:36 


624 


389376 


242970624 


24.9799920 


8.5453173 


.001(3025(34 


625 


390625 


244140625 


25.0000000 


8.5498797 


.001(500000 


62() 


391876 


245314376 


25.0199920 


8.5544372 


.001597444 


627 


393129 


246491883 


25.0399681 


8.5589899 


.00159489(5 


628 


394384 


247673152 


25.0599282 


8.5(535377 


.001592:357 


629 


395641 


248858189 


25.0798724 


8.5680807 


.001589825 


630 


smm 


250047000 


25.0998008 


8.5726189 


.001587:302 


631 


398161 


251239591 


25.1197134 


8.5771523 


.00158478(5 


632 


399424 


2524359()8 


25.1396102 


8.581(3809 


.001582278 


633 


400689 


253636137 


25.1594913 


8.58(52047 


.001579779 


()34 


401956 


254840104 


25.17935(36 


8.5907238 


.001577287 


635 


403225 


256047875 


25.19920(53 


8.5952380 


.00157480:5 


636 


404496 


257259456 


25.2190404 


8.5997476 


.001572:527 


637 


405769 


258474853 


25.2388589 


8.6042525 


.00 !.")( 59859 


638 


407044 


259694072 


25.2586(519 


8.608752(5 


.0015(57:39S 


639 


408321 


260917119 


25.2784493 


8.6132480 


.0015(34945 


640 


409600 


262144000 


25.2982213 


8.6177388 


.(X)15(52500 


641 


410881 


263374721 


25.3179778 


8.(5222248 


.0015600(52 


642 


412164 


264609288 


25.3377189 


8.(52(570(53 


.001557(5:32 


643 


41.3449 


265847707 


25.3.-.74447 


8.(5311830 


.001555210 


(^t4 


4147:3(^ 


267089984 


25..3771551 


8.(535(5551 


.0015527i)5 


645 


416025 


268336125 


25..3iH5S5()2 


8.(340122(3 


.001550:388 


646 


417316 


26958()136 


25.41(5."k>()1 


8.(544.")8.'')5 


.001547988 


647 


418009 


270840023 


25.43(51947 


8.(541K)437 


.001545595 


648 


419904 


272097792 


25.4.V)S441 


8.r).^);U974 


.00154:3210 


649 


421201 


273359449 


25.4754784 


8.(557941.5 


.0015408:32 



372 SQUARES, CUBES, SQUARE ROOTS, ETC. 



14 



SQUARES, CUBES 


, SQUARE ROOTS, 


CUBE 




ROOTS, AND RECIPROCALS 




Xo. 


Squares 


Cubes 


Square Roots 


Cube Roots 


Reciprocals 


650 


422500 


274625000 


25.4950976 


8.6623911 


.001538462 


651 


423801 


275894451 


25.5147016 


8.6()()8310 


.00153()098 


652 


425104 


277167808 


25.5342907 


8.6712665 


.001533742 


653 


426409 


278445077 


25.5538647 


8.6756974 


.001531394 


654 


427716 


279726264 


25.5734237 


8.6801237 


.001529052 


655 


429025 


281011375 


25.5929678 


8.6845456 


.001526718 


656 


430336 


282300416 


25.6124969 


8.6889()30 


.001524390 


657 


431649 


283593393 


25.6320112 


8.6933759 


.001522070 


658 


432964 


284890312 


25.6515107 


8.6977843 


.001519757 


659 


434281 


286191179 


25.6709953 


8.7021882 


.001517451 


660 


435G0O 


287496000 


25.6904652 


8.7065877 


.001515152 


661 


436921 


288804781 


25.7099203 


8.7109827 


.001512859 


662 


438244 


290117528 


25.7293607 


8.7153734 


.001510574 


663 


439569 


291434247 


25.7487864 


8.7197596 


.001508296 


664 


440896 


292754944 


25.7681975 


8.7241414 


.00150()024 


665 


442225 


294079625 


25.7875939 


8.7285187 


.001503759 


666 


443556 


295408296 


25.8069758 


8.7328918 


.001501502 


667 


444889 


296740963 


25.8263431 


8.7372604 


.001499250 


668 


446224 


298077632 


25.8456960 


8.7416246 


.001497006 


669 


447561 


299418309 


25.8650343 


8.7459846 


.001494768 


670 


448900 


300763000 


25.8843582 


8.7503401 


.001492537 


671 


450241 


302111711 


25.9036677 


8.7546913 


.001490313 


672 


451584 


303464448 


25.9229628 


8.7590383 


.001488095 


673 


452929 


304821217 


25.9422435 


8.7633809 


.001485884 


674 


454276 


306182024 


25.9615100 


8.7677192 


.001483680 


675 


455625 


307546875 


25.9807621 


8.7720532 


.001481481 


676 


456976 


308915776 


26.0000000 


8.7763830 


.0014792^)0 


677 


458329 


310288733 


26.0192237 


8.7807084 


.001477105 


678 


459684 


311665752 


26.0384331 


8.7850296 


.001474926 


679 


461041 


31304-6839 


26.0576284 


8.7893466 


.001472754 


680 


462400 


314432000 


26.0768096 


8.7936593 


.001470588 


681 


463761 


315821241 


26.0959767 


8.7979679 


.001468429 


682 


465124 


317214568 


26.1151297 


8.8022721 


.001466276 


683 


466489 


318611987 


26.1342687 


8.8065722 


.001464129 


684 


467856 


320013504 


26.1533937 


8.8108681 


.001461988 


685 


469225 


321419125 


26.172.-047 


8.8151598 


.001459854 


686 


470596 


322828856 


26.1916017 


8.8194474 


.001457726 


687 


471969 


324242703 


26.2106848 


8.8237307 


.001455604 


688 


473:^44 


325660672 


26.2297541 


8.8280099 


.001453488 


689 


474721 


327082769 


26.2488095 


8.8322850 


.001451379 


690 


476100 


328509000 


26.2678511 


8.8365559 


.001449275 


691 


477481 


329939371 


26.2868789 


8.8408227 


.001447178 


692 


478864 


331373888 


26.3058929 


8.8450854 


.001445087 


693 


480249 


332812557 


26.3248932 


8.8493440 


.001443001 


694 


481636 


334255384 


26.3438797 


8.8535985 


.001440922 


695 


483025 


335702375 


26.3628527 


8.8578489 


.001438849 


696 


484416 


33715:^536 


26.3818119 


8.8(;20952 


.00143()782 


697 


485809 


338608873 


26.400757() 


8.8()63375 


.001434720 


698 


487204 


34()0(;8392 


26.419()896 


8.8705757 


.001432(;()5 


699 


488(501 


3415320i)9 


26.4386081 


8.8748099 


.001430615 



SQUARES, CUBES, SQUARE ROOTS, ETC. 

15 



373 



SQUARES, CUBES 


, SQUARE ROOTS, 


CUBE 




ROOTS, AND RECIPROCALS 




No. 


Squares 


Cubes 


Square Roots 


Cube Roots 


Reciprocals 
.001428571 


700 


490000 


343000000 


20.45751:51 


8.8790400 


701 


4!)1401 


344472101 


2(5.47(5404(5 


8.88:52(561 


.0014205:54 


702 


492804 


^45948408 


26.4952826 


8.8874882 


.001424.-)01 


703 


494209 


347428927 


26.5141472 


8.89170(53 


.001422475 


704 


495(}16 


348913(5(54 


26.5329983 


8.895il2(J4 


.001420455 


705 


497025 


350402(525 


26.5518:5(51 


8.9001:504 


.(H)1418440 


706 


49843() 


351895810 


2(5.570(5(505 


8.iM)4:5:5(5() 


.00141(5431 


707 


499849 


353393243 


2(5.5894710 


8.9085:587 


.001414427 


708 


501204 


354894912 


26.(5082(594 


8.9127:5(59 


.001412429 


709 


502081 


350400829 


20.0270539 


8.9169311 


.001410437 


710 


504100 


357911000 


26.6458252 


8.9211214 


.001408451 


711 


505521 


359425431 


26.6(5458:53 


8.925:5078 


.00140(5470 


712 


50(5944 


3(50944128 


26.(58:5:5281 


8.9294902 


.001404494 


713 


5083(59 


3024(57097 


2(5.7020598 


8.9:3:5(5(587 


.001402525 


7U 


509790 


3(515994344 


26.7207784 


8.9:5784:53 


.0014(XJ5(50 


715 


511225 


305525875 


2(5.73948:59 


8.9420140 


.001:598(501 


710 


512(356 


3670(51(590 


1:6.75817(53 


8.94(51809 


.001:59(5(548 


717 


514089 


308001813 


26.7768557 


8.9503438 


.001:594700 


718 


515524 


37014(5232 


26.7955220 


8.9545029 


.a)i:592758 


719 


510901 


371094959 


20.8141754 


8.958(5581 


.001390821 


720 


518400 


373248000 


26.8328157 


8.9(528095 


.001:588889 


721 


519841 


3748053(51 


26.85144:52 


8. i)( 5(59570 


.OOi:58(51K53 


722 


521284 


3703(57048 


26.8700577 


8.9711007 


.001:585042 


723 


522729 


37793:50(57 


20.888(5593 


8.9752406 


.001383126 


724 


524170 


379503424 


26.9072481 


8.979:57(56 


.001:581215 


725 


525(525 


381078125 


26.9258240 


8.98:55089 


.001:579310 


72(5 


527070 


382(557170 


20.i)44:5872 


8.987(5:573 


.001:577410 


727 


528529 


384240583 


2(5.9(52<):;75 


8.9917(520 


.001:575516 


728 


529984 


385828352 


20.9814751 


8.9958829 


.001:57:5(526 


729 


531441 


387420489 


27.0000000 


9.0000000 


.001371742 


730 


532900 


389017000 


27.0185122 


9.00411:54 


.001:5(598(53 


731 


5;U3(;i 


390617891 


27.0:570117 


9.008222i) 


.001:5(57989 


732 


535824 


392223108 


27.0554985 


9.0123288 


.001:5(5(5120 


733 


537289 


393832837 


27.07:59727 


9.01(54309 


.001:5(5425(5 


lU 


53875() 


39544(5904 


27.0924:U4 


9.0205293 


.001:5(52:598 


735 


540225 


3970(55375 


27.11088:54 


9.024(52:59 


.001:5(50.544 


73() 


541(59(5 


398(588256 


27.1293199 


9.0287149 


.0()i:558()^H) 


737 


5431(59 


400315553 


27.14774:59 


9.0:528021 


.00i:55(58,"')2 


738 


544(^44 


401947272 


27.1(5(51554 


9.0:5(58857 


.001:555014 


739 


540121 


40358:1419 


27.1845544 


9.0401H)55 


.00K553180 


740 


547(500 


405224000 


27.202!)410 


9.0450417 


.001:551:5,51 


741 


549081 


40(58(59021 


27.2213152 


9.0491142 


.001:549528 


742 


5505(54 


408518488 


27. 2:5<H 57(59 


9.05318:51 


.(.H)i:547709 


743 


552049 


410172407 


27.25802(5:5 


9.0572482 


.001:545895 


744 


553530 


4118:50784 


27.27(5:5(5:U 


9.0(51 :U)98 


.00i:54408(> 


745 


555025 


41:549:5(525 


27. 2! M( 5881 


9.(X 55:5(577 


.001:542282 


74() 


55(5510 


4151 (509:5(5 


27.31:5000(5 


9. (M 594220 


.001:540483 


747 


558009 


41(58:52723 


27.:53i:5(M)7 


9.07:5472(5 


.001:5:58(588 


748 


559504 


418508992 


27.:549.~)887 


9.0775197 


.001:5:5(5898 


749 


501001 


420189749 


27.:5(578()44 


- 9.(3815(5:51 , 


.001:5:55113 



374 SQUARES, CUBES, SQUARE ROOTS, ETC. 

16 



X 



SQUARES, CUBES, 


SQUARE ROOTS, 


CUBE 




ROOTS, AND RECIPROCALS 




JVo. 


Squares 


Cubes 


Square Roots 


Cube Roots 


Reciprocals 


7r.o 


562500 


421875000 


27.3861279 


9.0856030 


.001333333 


751 


564001 


423564751 


27.4043792 


9.0896392 


.001331558 


752 


565504 


425259008 


27.4226184 


9.0936719 


.001329787 


753 


567009 


426957777 


27.4408455 


9.0977010 


.001328021 


754 


568516 


428661064 


27.4590604 


9.1017265 


.001326260 


755 


570025 


430368875 


27.4772633 


9.1057485 


.001324503 


756 


571536 


432081216 


27.4954542 


9.1097669 


.001322751 


757 


573049 


433798093 


27.5136330 


9.1137818 


.001321004 


758 


574564 


435519512 


27.5317998 


9.1177931 


.001319261 


759 


576081 


437245479 


27.5499546 


9.1218010 


.001317523 


760 


577600 


438976000 


27.5680975 


9.1258053 


.001315789 


761 


579121 


440711081 


27.5862284 


9.1298061 


.001314060 


762 


580644 


442450728 


27.6043475 


9.1338034 


.001312336 


7(53 


582169 


444194947 


27.6224546 


9.1377971 


.001310616 


764 


583696 


445943744 


27.6405499 


9.1417874 


.001308901 


765 


585225 


447697125 


27.6586334 


9.1457742 


.001307190 


766 


586756 


449455096 


27.6767050 


9.1497576 


.001305483 


767 


588289 


451217663 


27.6947648 


9.1537375 


.001303781 


768 


589824 


452984832 


27.7128129 


9.1577139 


.001302083 


769 


591361 


454756609 


27.7308492 


9.1616869 


.001300390 


770 


592900 


456533000 


27.7488739 


9.1656565 


.001298701 


771 


594441 


458314011 


27.7668868 


9.1696225 


.001297017 


772 


595984 


4(J0099()48 


27.7848880 


9.1735852 


.001295337 


773 


597529 


461889917 


27.8028775 


9.1775445 


.001293661 


774 


599076 


463684824 


27.8208555 


9.1815003 


.001291990 


775 


600025 


465484375 


27.8388218 


9.1854527 


.001290323 


776 


602176 


467288576 


27.8567766 


9.1894018 


.001288660 


777 


603729 


469097433 


27.8747197 


9.1933474 


.001287001 


778 


605284 


470910952 


27.8926514 


9.1972897 


.001285347 


779 


60()841 


472729139 


27.9105715 


9.2012286 


.001283697 


780 


608400 


474552000 


27.9284801 


9.2051641 


.001282051 


781 


601M!61 


476379541 


27.94()3772 


9.2090962 


.001280410 


782 


611524 


478211768 


27.9()42()29 


9.2130250 


.001278772 


783 


613089 


480048687 


27.9821372 


9.2169505 


.001277139 


784 


614656 


481890304 


28.0000000 


9.2208726 


.001275510 


785 


616225 


4837:5()(;25 


28.0178515 


9.2247914 


.001273885 


786 


617796 


485587656 


28.0356915 


9.2287068 


.0012722(55 


787 


619369 


487443403 


28.0535203 


9.2326189 


.001270648 


788 


620944 


489303872 


28.0713377 


9.2365277 


.00126^)036 


789 


622521 


491169069 


28.0891438 


9.2404333 


.001267427 


790 


624100 


493039000 


28.1069386 


9.2443355 


.001265823 


791 


625681 


494913()71 


28.1247222 


9.2482344 


.001264223 


792 


6272f)4 


49()793088 


28.1424946 


9.2521300 


.001262626 


793 


()28849 


498()77257 


28.1602,557 


9.2560224 


.001261034 


794 


6304:^.6 


50056()184 


28.17800.56 


9.2599114 


.001259446 


795 


632025 


502459875 


28.1957444 


9.2637973 


.001257862 


79f) 


633()16 


504358^,36 


28.2134720 


6.2(576798 


.001256281 


797 


635209 


50()261573 


28.2311884 


9.2715592 


.001254705 


798 


63>6804 


5081()9592 


28.2488938 


9.27.54352 


.001253133 


799 


638401 


510082399 


28.2()65881 


9.2793081 


.001251564 



SQUARES, CUBES, SQUARE ROOTS, ETC, 

17 



(O 



SQUARES, CUBES, SQUARE ROOTS, CUBE 
ROOTS, AND RECIPROCALS 



Ko. 


Squares 


Cubes 


Square Roots 


Cube Roots 


Reciprocals 


800 


()40000 


512000000 


28.2842712 


9.2831777 


.0()i2r)()(X)() 


801 


641()()1 


513922401 


28.:5019434 


9.2870440 


.00124s };;<) 


802 


()43204 


515849()08 


28.31 9()045 


9.2i)0i)072 


.0O124(5S,S3 


803 


644809 


517781627 


28.:5:;7254(j 


9.2947(571 


.001245:5:50 


804 


646416 


51971S464 


28.:55489:58 


9.298(52::59 


.00124:5781 


805 


648025 


5216()()125 


28.:3725219 


9.3024775 


.0012422:5(5 


806 


649()36 


523()0()()16 


28.390L391 


9.30(5:5278 


.001240(595 


807 


651249 


525557^)43 


28.4077454 


9.3101750 


.001 2:59 1.-)7 


808 


652864 


527514112 


28.425:5408 


9.:3140190 


.0012:57(524 


809 


654481 


529475129 


28.4429253 


9.3178599 


.0012:5(5094 


810 


656100 


531441000 


28.4(;04989 


9.321(5975 


.0012:345(58 


811 


657721 


533411731 


28.4780(517 


9.3255:520 


.0012:5:504(5 


812 


659344 


535387328 


28.4i)5()i:57 


9.;329:5(5:34 


.001231.V27 


813 


660969 


5373()7797 


28.5131549 


9.3:331916 


.0O12:;00I2 


814 


6625i)6 


539353144 


28.5:50(5852 


9.:3:5701(57 


.001228501 


815 


6(J4225 


54134:5375 


28.5482048 


9.:5408:586 


.00122(5994 


816 


&mm 


54333849() 


28.5(>57L37 


9.344(5575 


.00122541K) 


817 


667489 


54533S513 


28.58:52119 


9.3484731 


.00122:^990 


818 


669124 


54734:^132 


28. 6001)99:5 


9.3522857 


.001222494 


819 


670761 


549:353259 


28.61817(50 


9.3560952 


.001221001 


820 


672400 


55i:')()S000 


28.():55i)421 


9.3599016 


.001219512 


821 


674041 


55:3;)87661 


28.65:5097(5 


9.:36:57049 


.00121.S()27 


822 


675()84 


555412248 


28.(^705424 


9.:^(575051 


.00121(5.")45 


823 


677329 


557441767 


28.6S797()6 


9.:37i:^022 


.00121.-.0(57 


824 


678976 


559476224 


28.7054002 


9.:^750iK)3 


.00121:5592 


825 


680625 


561515625 


28.72281:^2 


9.: 5788873 


.001212121 


826 


682276 


5(i:;55997l> 


28.7402157 


9.:5826752 


.001210(5.")4 


827 


683i)29 


5(i5()09283 


28.757(5077 


9.:5864(;00 


.001209190 


823 


()8r)584 


5()7()'>:r)52 


28.7749891 


9.:5902419 


.001207729 


829 


687241 


569722789 


28.792:5(501 


9.3940206 


.00120(5273 


830 


688900 


571787000 


28.809720(5 


9.:59779(54 


.001204819 


831 


6905()1 


57:^856191 


28.827070(5 


9.4015(591 


.00120:5:^(59 


832 


692224 


5759:](K3()8 


28.8444102 


9.405:5:587 


.001201023 


833 


693889 


5780095:^7 


28.8(517:594 


9.4091054 


.0012OO4S0 


834 


695556 


58009:5704 


28.8790582 


9.4128()t)0 


.0011!M)041 


835 


697225 


582182875 


28.8<K5:5(5(;6 


9.4! 0(5297 


.001197(505 


8;3() 


6i)8896 


584277056 


28.91:5(5(54(5 


9.420:5873 


.001H)(5172 


837 


70()5()9 


58():i7()253 


28.9:50952:5 


9 4241420 


.(H-) 11^743 


838 


702244 


588480472 


28.9482297 


9.42789:56 


.(X) 119:5317 


839 


703921 


590589719 


28.^)(554^K57 


9.4:51(5423 


.001191895 


840 


705()00 


592704000 


28.98275.35 


9.4:-5:5880 


.0011W476 


841 


707281 


594.s2;5:;2l 


29.0000000 


94:591:507 


.00118<)061 


842 


70S9()4 


59(;947688 


29.0 172; 5(5: 5 


9.4428704 i 


.001187(548 


843 


710()49 


599077107 


29.0:544(523 


9.44(1(1072 i 


.00118(5240 


844 


71233() 


()()12115S4 


29 051(5781 


9.450:5410 1 


.ooiis4.s:u 


845 


714025 


(;o:;:;5ii25 


29()i;s88:57 


9.4540719 1 


.00118:^4:52 


846 


71571() 


()0549:)7:5<i 


29 (1S( 50791 


9.4577! »99 


.0011820:33 


847 


717409 1 


(;()7t;4:)423 


29.10:52(544 


9. 4(51. "5249 


.001180(5:58 


848 


719104 


609800192 


29.120i:5iK5 


9.4(5.")2470 


.001179245 


849 


720801 


611960049 


29.137(504(5 . 


9.4()8<)(5(51 


.001177856 



376 SQUARES, CUBES, SQUARE ROOTS, ETC. 

18 



SQUARES, CUBES 


, SQUARE ROOTS, 


CUBE 




ROOTS, AND RECIPROCALS 




Jfo. 


Squares 


Cubes 


Square Roots 


Cube Roots 


Reciprocals 


850 


722500 


614125000 


29.1547595 


9.4726824 


.001176471 


851 


724201 


616295051 


29.1719043 


9.47(53957 


.001175088 


852 


725904 


618470208 


29.1890390 


9.48010(51 


.001173709 


853 


727609 


620650477 


29.2061(537 


9.4838136 


.001172333 


854 


729316 


622835864 


29.2232784 


9.4875182 


.0011709(50 


855 


731025 


62502(J375 


29.2403830 


9.4912200 


.00116^*591 


856 


732736 


627222016 


29.2574777 


9.4949188 


.001168224 


857 


734449 


629422793 


29.2745623 


9.4986147 


.0011(5(5861 


858 


736164 


631628712 


29.291(5370 


9.5023078 


.001165501 


859 


737881 


633839779 


29.3087018 


9.5059980 


.001164144 


860 


739600 


636056000 


29.3257566 


9.5096854 


.001162791 


861 


741321 


638277381 


29.3428015 


9.5133699 


.001161440 


862 


743044 


640503928 


29.3598365 


9.5170515 


.001160093 


863 


744769 


642735647 


29.3768(516 


9.5207303 


.001158749 


8()4 


746496 


644972544 


29.3938769 


9.5244063 


.001157407 


865 


748225 


647214625 


29.4108823 


9.5280794 


.00115(50(59 


866 


749956 


649461896 


29.4278779 


9.5317497 


.001154734 


867 


751689 


651714363 


29.4448637 


9.5354172 


.001153403 


868 


753424 


653972032 


29.4618397 


9.5390818 


.001152074 


869 


755161 


656234909 


29.4788059 


9.5427437 


.001150748 


870 


756900 


658503000 


29.4957624 


9.5464027 


.001149425 


871 


758641 


660776311 


29.5127091 


9.5500589 


.001148106 


872 


760384 


663054848 


29.52964(51 


9.5537123 


.001146789 


873 


762129 


665338617 


29.54(55734 


9.5573(530 


.001145475 


874 


763876 


667627()24 


29.5634910 


9.5610108 


.001144165 


875 


765625 


669921875 


29.5803989 


9.564(5559 


.001142857 


876 


767376 


672221376 


29.5972972 


9.5(582982 


.001141553 


877 


769129 


674526133 


29.6141858 


9.5719377 


.001140251 


878 


770884 


676836152 


29.(5310648 


9.5755745 


.001138952 


879 


772641 


679151439 


29.6479342 


9.5792085 


.001137656 


880 


774400 


681472000 


29.6647939 


9.5828397 


.001136364 


881 


77()161 


683797841 


29.6816442 


9.58(54(582 


.001135074 


882 


777924 


686128968 


29.6984848 


9.5900939 


.001133787 


883 


779689 


688465oS7 


29.7153159 


9.59371(59 


.001132503 


884 


781456 


690807104 


29.7321375 


9.5973373 


.001131222 


885 


78:5225 


693154125 


29.7489496 


9.6009548 


.001129944 


886 


784996 


69550()45() 


29.7(557521 


9.(5045696 


.001128668 


887 


786769 


697864103 


29.7825452 


9.(5081817 


.001127396 


888 


788544 


700227072 


29.7993289 


9.(5117911 


.001126126 


889 


790321 


7025953()9 


29.81(51030 


9.6153977 


.001124859 


8^K) 


792100 


7049()9000 


29.8328678 


9.6190017 


.001123596 


891 


793881 


707347971 


29.849(5231 


9.(522(5030 


.001122334 


892 


7956()4 


709732288 


29.8(5(53690 


9.62(52016 


.001121076 


893 


797449 


712121957 


29.8831056 


9.6297975 


.001119821 


894 


799236 


714516984 


29.81)98328 


9.6333907 


.0011185(58 


895 


801025 


716917375 


29.91(55506 


9.(53(59812 


.001117318 


896 


802S1() 


71932:U36 


29.9332591 


9.(5405(590 


.00111(5071 


897 


804609 


721734273 


29.9499583 


9.(5441542 


.001114827 


898 


806404 


724150792 


29.9(5(5(5481 


9.6477367 


.001113586 


8i^9 


808201 


72(5572(599 


29.9833287 


9.(5513166 


.001112347 



SQUAIiES, CUBES, SQUARE BOOTS, ETC. 

19 



377 



SQUARES, CUBES, SQUARE ROOTS, 


CUBE 




ROOTS, AND RECIPROCALS 




IVo. 


Squares 


Cal)es 


Square Roots 


Cube Roots 


Rftcilirocals 


900 


810000 


729000000 


30.0000000 


9.(55489:38 


.001111111 


901 


811801 


731432701 


30.016(5(520 


9.(S5S4(),S4 


.001109878 


902 


813004 


733870808 


30.0:533148 


9.(5(520403 


.001I()8()47 


1X)3 


815409 


736314327 


30.0499584 


9.(5(55(5(KK) 


.001107420 


iH)4 


817210 


7387632()4 


30.06(55928 


• 9.(5(5917(52 


.001106195 


905 


81<K)25 


741217625 


30.08:32179 


9.(5727403 


.001104972 


900 


820836 


743()77416 


30.09i)83:39 


9.(57(53017 


.00110:5753 


907 


822649 


74() 142(543 


30.11(54407 


9.(5798(504 


.0011025:5(5 


908 


8244()4 


748613312 


30.i:i30383 


9.(58:541(56 


.00110L322 


909 


826281 


751089429 


30.1496269 


9.68(59701 


.001100110 


910 


828100 


753571000 


30.1(562063 


9.(5905211 


.001098901 


911 


829921 


7.~6058()31 


30.1827765 


9.(5940(594 


.001097(595 


912 


831744 


758550528 


30.1993377 


9.(597(5151 


.00109(5491 


913 


8335()9 


761048497 


30.2158899 


9.7011583 


.()0109.V2i)0 


914 


83539() 


763551944 


30.2324:529 


9.704()<'89 


.001094092 


915 


837225 


7(56060875 


30.2489(5(59 


9.7082:5(59 


.001092896 


91 G 


839()5() 


768575296 


30.2(554919 


9.7117723 


.001091703 


917 


840889 


771095213 


30.2820079 


9.715:5051 


.001090513 


918 


842724 


773620(532 


30.2985148 


9.7188:554 


.001089325 


919 


844561 


776151559 


30.3150128 


9.7223631 


.0010881:59 


920 


846400 


778688000 


30.;5315018 


9.7258883 


.00108(5957 


921 


848241 


781229<h;1 


30.:5479818 


9.7294109 


.001085776 


922 


850084 


783777448 


30.:i644529 


9.7329:509 


.001084r)i»9 


923 


851929 


78()3;m()7 


30.3809151 


9.7:;(54484 


.00108:5424 


924 


85;'>776 


788881K)24 


30.397:5(583 


9.7:>99(534 


.0010.S2251 


925 


855(525 


791453125 


30.41:58127 


9.74:54758 


.001 (IS 1 081 


926 


85747() 


794022776 


30.4:502481 


9.74(59857 


.001079914 


927 


8593,29 


71K)597983 


30.44(5(5747 


9.75049:50 


.001078749 


928 


861184 


799178752 


30.4(5:50924 


9.75:59979 


.00107758(5 


929 


8():i041 


80176r;089 


30.4795013 


9.7575002 


.00107(5426 


930 


8(54900 


804357000 


30.4959014 


9.7610001 


.(X)10752(59 


931 


8()()761 


806954491 


30.5122926 


9.7(544974 


.001074114 


932 


8()8624 


8095575(58 


30.5286750 


9.7(579922 


.()O1072!i(51 


fl33 


870489 


8121(5()237 


30.5450487 


9.7714845 


.001071811 


934 


872;}5() 


814780504 


30.5(5141:56 


9.7749743 


.001070(5(54 


93,5 


874225 


817400375 


30.5777(597 


9.7784(516 


.001()(5<)519 


93(3 


87(;09() 


820025856 


30.5941171 


9.78194(56 


.0010(58:576 


937 


877969 


822(55(5953 


30.6104557 


9.7854288 


.0010(572:56 


938 


879844 


825293(572 


30.62(57857 


9.788<)()87 


.0010(5(5098 


939 


881721 


82793(5019 


30.(54:51069 


9.7il2:5861 


.0010(549(53 


940 


883600 


830584000 


:50.6594194 


9.7958611 


.0010(5:58:50 


941 


885481 


8:5323.7(521 


:30.(-;757233 


9.799:5:5:5() 


.O01(^(52()<><) 


942 


8873()4 


83,589(5888 


30.(5920185 


9.802S0:5() 


.0010(51571 


943 


8S9249 


8385(51807 


30.708:5051 


9.80(52711 


.0010(50445 


944 


891 13() 


8412:523,84 


30.72458:50 


9.8(V.)7:5(52 


.0010.'')9:522 


945 


893025 


84:5908(525 


30.740S.V2;) 


9.8131989 


.(HI 1058201 


940 


894911; 


84( 5590.1:56 


:50.757li:50 


9.81(5(5591 


.(M)10.")7082 


947 


89()S09 


849278123 


:50.77:5:5(551 


9.S2011()9 


.0010559(5(5 


948 


898704 


851971:592 


30.789(5086 


9.82:5.-)72:i 


.001054S52 


949 


900601 


854(570:549 


30.80584:56 


9.8270252 


.00105:5741 



378 SQUARES, CUBES, SQUARE ROOTS, ETC. 

20 



SQUARES, CUBES 


, SQUARE ROOTS, 


CUBE 




ROOTS, AND RECIPROCALS 




Jfo. 


Squares 


Cubes 


Square Roots 


Cube Roots 


Reciprocals 


950 


902500 


857375000 


30.8220700 


9.8304757 


.001052632 


951 


904401 


860085351 


30.8382879 


9.8339288 


.001051525 


952 


906304 


862801408 


30.8544972 


9.8373695 


.001050420 


953 


908209 


865523177 


30.8706981 


9.8408127 


.001049318 


954 


910116 


868250664 


30.8868904 


9.8442536 


.001048218 


955 


912025 


870983875 


30.9030743 


9.8476920 


.001047120 


956 


913936 


873722816 


30.9192497 


9.8511280 


.00104(5025 


957 


915849 


876467493 


30.9354166 


9.8545617 


.001044932 


958 


917764 


879217912 


30.9515751 


9.8579929 


.001043841 


959 


919681 


881974079 


30.9677251 


9.8614218 


.001042753 


960 


921600 


884736000 


30.9838668 


9.8648483 


.001041667 


961 


923521 


887503681 


31.0000000 


9.8682724 


.001040583 


962 


925444 


890277128 


31.0161248 


9.8716941 


.001039501 


963 


927369 


893056347 


31.0322413 


9.8751135 


.001038422 


964 


929296 


895841344 


31.0483494 


9.8785305 


.001037344 


965 


931225 


898632125 


31.0644491 


9.8819451 


.001036269 


966 


933156 


901428(596 


31.0805405 


9.8853574 


.001035197 


967 


935089 


904231063 


31.0966236 


9.8887673 


.001034126 


968 


937024 


907039232 


31.1126984 


9.8921749 


.001033058 


969 


938961 


909853209 


31.1287648 


9.8955801 


.001031992 


970 


940900 


912673000 


31 .1448230 


9.8989830 


.001030928 


971 


942841 


915498611 


31.1608729 


9.9023835 


.001029866 


972 


944784 


918330048 


31.1769145 


9.9057817 


.001028807 


973 


946729 


921167317 


31.1929479 


9.9091776 


.001027749 


974 


948676 


924010124 


31.2089731 


9.9125712 


.001026694 


975 


950625 


926859375 


31.2249900 


9.9159(524 


.001025(541 


976 


952576 


929714176 


31.2409987 


9.9193513 


.001024590 


977 


954529 


932574833 


31.2569992 


9.9227379 


.001023541 


978 


956484 


935441352 


31.2729915 


9.9261222 


.001022495 


979 


958441 


938313739 


31.2889757 


9.9295042 


.001021450 


980 


960400 


941192000 


31.3049517 


9.9328839 


.001020408 


981 


962361 


944076141 


31.3209195 


9.9362613 


.0010193(58 


982 


964324 


94696()168 


31.33(38792 


9.9396363 


.001018330 


983 


96()289 


949862087 


31.3528308 


9.9430092 


.001017294 


984 


968256 


952763<)04 


31.3687743 


9.94(i3797 


.00101()2(50 


985 


970225 


955(571625 


31.3847097 


9.9497479 


.001015228 


986 


972196 


958585256 


3i.400(;;i(;9 


9.95:51138 


.001014199 


987 


9741()9 


961504803 


31.41()5561 


9.9;5()4775 


.001013171 


988 


976144 


964430272 


31. 43241 ;73 


9.9598389 


.001012146 


989 


978121 


967361()69 


31.4483704 


9.9(531981 


.001011122 


990 


980100 


970299000 


31.4f)42654 


9.9665549 


.001010101 


991 


982081 


973242271 


31.4801525 


9.9(;9{K)95 


.0010(^9082 


9i)2 


984064 


976191488 


31.49(50315 


9.9732()19 


.0010080()5 


993 


98(5019 


97914()()57 


31.5119025 


9.9766120 


.001007049 


994 


988036 


982107784 


31.5277(555 


9.9799599 


.001006036 


995 


99(J()25 


985074875 


31.5436206 


9.983;'.055 


.001005025 


996 


992016 


988047936 


31.5."')9-1()77 


9.98(5(5488 


.001004016 


997 


994009 


9i)102()973 


3i.575:;()()8 


9.9899<)00 


.()()10();5009 


998 


99(;001 


99401 19i)2 


31.5911380 


i). 9933289 


.001002004 


999 


998001 


997(X)2999 


31.(5(J(5i)(513 


i).99(;()(55() 


.001001001 



APPENDIX V 
CONVERSION TABLES 



CONVERSION TABLES 



381 



TABLES 


FOR CONVERTING UNITED STATES 






WEIGHTS AND MEASURES 








METRIC TO CUSTOMARY 








WEIGHTS 






Milli^n-iuiis 


(^raius 


Grams 


Kilograms 


Tonnes to 


Tonnes to 


No. 


to 


to 


to Avoirdii})ois 


to Avoirdupois 


Net Tons of 


(Jross Tons of 


1 


Grains 


Troy Ounces 


Ounces 


Pounds 


2000 Pounds 


2240 Pounds 


.01543 


.03215 


.03527 


2.20462 


1.10231 


.98421 


2 


.0308G 


.0()4;30 


.07055 


4.40924 


2.20462 


1.96841 


3 


.046:30 


.09645 


.10582 


6.()1387 


3.:5()693 


2.952(;2 


4 


.06173 


.12860 


.14110 


8.81849 


4.40924 


3.9:J682 


5 


.07716 


.1()075 


.176:57 


11.02311 


5.51156 


4.92103 


6 


.09259 


.192^)0 


.21164 


13.22773 


6.6i:387 


5.90.-)24 


7 


.10803 


.22.-)06 


.24()92 


15.4:52:36 


7.71618 


(5.88944 


8 


.121346 


.25721 


.28219 


17.63698 


8.81849 


7.87:565 


9 


.13889 


.2893(5 


.31747 


19.84160 


9.92080 


8.85785 






1 Kilogram = 1543 


2.356:39 Grains 








LINEAR Ml 


EASURE 






Miliimetrrs 


Centimeters 


Meters 


liletcrs 


Kilometers 


Kilometers 


No. 


to m\\s of ail 


to 


to 


to 


to 


to 




Inch 


Inches 


Feet 


Yards 


Statute Miles 


Nautical Miles 


1 


2.51968 


.39370 


3.280833 


1.093611 


.62137 


.53959 


2 


5.03936 


.78740 


e.oinmi 


2.187222 


1.24274 


1.07919 


3 


7.55904 


1.18110 


9.842.500 


3.2808:13 


1.86411 


1.(51878 


4 


10.07872 


1.57480 


13.12.33:53 


4.:i74444 


2.4S548 


2.158:57 


5 


12.59840 


1.96850 


16.4041()7 


5.468056 


3.10685 


2.6979(5 


6 


15.11808 


2.36220 


lO-fiS-^OOO 


6.561()(;7 


3.72822 


3.2:575(5 


7 


17.63776 


2.75.^)<>0 


22.96.~)S:;:i 


7.65.-)278 


4.:U959 


3.77715 


8 


20.15744 


3.141K)0 


26.24()(;67 


8.748S89 


4.97096 


4.:51(574 


9 


22.67712 


3.54330 


29.527500 


9.842500 


5.59233 


4.85633 


• 



382 



CONVERSION TABLES 



TABLES 


FOR CONVERTING UNITED STATES 






WEIGHTS AND 


MEASURES 








CUSTOMARY TO METRIC 










WEIGHTS 






Grains 


: Troy Ounces 


Avoirdupois 


Avoirdupois 


Net Tons of 


Gross Tons of 


No. 


to 


to 


Ounces 


Pounds to 


2000 Pounds 


2240 Pounds 




Milligrams 


Grams 


to Grams 


Kilograms 


to Tonnes 


to Tonnes 


1 


64.79892 


31.10348 


28.34953 


.45359 


.90718 


1.01605 


2 


129.59784 


62.20696 


56.69f)05 


.^)0718 


1.81437 


2.03209 


3 


194.39675 


93.31044 


85.04858 


1.36078 


2.72155 


3.04814 


4 


259.19567 


124.41392 


113.39811 


1.81437 


3.62874 


4.06419 


5 


323.99459 


155.51740 


141.74763 


2.26796 


4.53592 


5.08024 


6 


388.79351 


186.62088 


170.09716 


2.721.55 


5.44311 


6.09628 


7 


453.59243 


217.72437 


198.44669 


3.17515 


6.35029 


7.11233 


8 


518.39135 


248.82785 


226.79621 


3.62874 


7.25748 


8.12838 


9 


583.19026 


279.93133 


255.14574 


4.08233 


8.16466 


9.14442 






1 AvoirduiD 
L 


ois Pound = 
-INEAR ME 


453.5924277 Grams 
EASURE 






64tlis of an 


Inches 


Feet 


Yards 


Statute Miles 


Nautical Miles 


M 


Inch to 


to 


to 


to 


to 


to 




Millimeters 


Centimeters 


Meters 


Meters 


Kilometers 


Kilometers 


1 


.39688 


2.54001 


.304801 


.914402 


1.60935 


1.85325 


2 


.79375 


5.08001 


.609601 


1.828804 


3.21869 


3.70650 


3 


1.19063 


7.62002 


.914402 


2.743205 


4.82804 


5.55975 


4 


1.58750 


10.16002 


1.219202 


3.657607 


6.43739 


7.41300 


5 


1.98438 


12.70003 


1.524003 


4.572009 


8.04674 


9.26625 


6 


2.38125 


15.24003 


1.828804 


5.486411 


9.65608 


11.11950 


7 


2.77813 


17.78004 


2.133604 


6.400813 


11.26543 


12.97275 


8 


3.17501 


20.32004 


2.438405 


7.315215 


12.87478 


14.82600 


9 


3.57188 


22.86005 


2.743205 


8.229()16 


14.48412 


16.67925 






1 Nautic 


ialMile = 


1853.25 Meters 








1 Gunte 


r's Chain = 


20.1 1()8 Meters 








1 Fathoi 


m = 


1.829 Meters 





INDEX 



Absorption dynamometer, 305. 
Acceleration, 124, 125, 131, 143, 170, 
171, 172. 

angular, 170. 

normal, 144. 

taniijential, 144. 
Angular velocity, 169, 196. 
Appendix I, Hyperbolic Functions, 339. 
Appendix II, Logarithms of Numbers, 

345. 
Appendix III, Trigonometric Func- 
tions, 349. 
Appendix IV, Squares, Cubes, etc., 359. 
Appendix V, Conversion Tables, 380. 
Attractive force, 132, 136. 

Ball bearings, 280. 
Bearings, ball, 280. 

roller, 279. 
Belts, centrifugal tension, 293. 

coefficient of friction, 292. 

creeping of, 292. 

friction of, 288. 

stiffness of, 2i)4. 
Body, freely falling, 126. 

projected up inclined plane, 158. 

projected upward, 126. 

through atmosphere, motion of, 138. 
Brake friction, 306, 307. 
Brake shoes, friction of, 310. 
Brake shoe testing machine, 252. 

Car on single rail, 227. 

Catenary, 118. 

Center of gravity, 27, 32. 

of cone, 33. 

of locomotive counterbalance, 40. 

of rail section, 46. 

of triangle, 35. 

of T-section, 30. 

of U-section, 30. 
Center of percussion, 187, 330. 



Centrifugal force, 146. 
Centrifugal tension of belts, 293. 
Circular pendulum, 148. 
Coefficient of friction, 261. 
Combined rotation and translation 

173. 
Compound pendulum, 188. 
Concurrent forces, 5. 

in plane, 9. 

in space, 14. 
Conical pivot, 301. 
Connecting rod, 212. 
Conservation of energy, 234. 
Conversion Tables, 381. 
Cords, and pulleys, 113. 

flexible, 111. 

uniform load along cord, 117. 

uniform load horizontally, 114. 
Couples, 50, 53, 54. 
Creeping of belts, 292. 
Cubes, Cube Roots, etc., 359. 
Curvilinear motion, 142. 
Cycloidal pendulum, 154. 

D'Alembert's principle, 177. 
Determination of //, 192. 
Direct central impact, 316, 319. 
Direct eccentric impact, 328. 
Displacement, 4. 
Dry surfaces, friction of, 262. 
Durand's rule, 46. 
Dynamometer, absorption, 305. 
transmission, 291. 

Eccentric impact, 328. 
Elasticity of materials, 322. 
Ellipse of inertia, 92. 
Ellipsoid of inertia, 105. 
Energy, 2:i3. 

and work, 229. 

conservation of, 234. 

of body moving in straight line, 234. 



383 



38-i 



INDEX 



Experimental determination of mo- 
ment of inertia, 191. 

Falling bodies, 126. 
Flat pivot, 299. 
Flexible cords, 111. 
Force, 1, 8, 56, 63. 

moment of, 17, 18. 

parallel, 20, 22. 

polygon of, 7. 

representation of, 5. 

tangential and normal, 146. 

transmissibility of, 8. 

triangle of, 6. 

units of, 1. 
Friction, 261. 

coefficient of, 261. 

laws of, dry surfaces, 262. 

laws of, lubricated surfaces, 264. 

of belts, 288, 292. 

of brake shoes, 310. 

of pivots, 299. 

of worn bearing, 297. 

rolling, 272. 
Friction brake, 306, 307. 
Friction gears, 285. 
Friction wheels, 274. 

Gears, friction, 285. 
Gravity, center of (see Center of grav- 
ity). 
Gyroscope, 216. 
Gyroscopic action explained, 221. 

Harmonic motion, 132. 
Hyperbolic Functions, 119, 339. 

Impact, 315. 

direct central, elastic, 319. 

direct central, inelastic, 316. 

imperfectly elastic bodies, 323. 

oblique, 331. 

rotating bodies, 332. 

tension and compression, 325. 
Inclined plane, motion on, 128. 
Inertia, 2. 

ellipse of, 92. 

ellipsoid of, 105. 

moment of, 69, 71. 

(see Moment of inertia). 
Introduction, 1. 



Kinetic energy of rolling bodies, 256. 

Laws of friction, 262, 264. 

of motion, 127. 
Length of cord, 116, 122. 
Locomotive counterbalance, 40. 
Locomotive side rod, 211. 
Logarithms of Numbers, 345. 
Lubricants, testing of, 270. 
Lubricated surfaces, friction of, 264. 

Mass, 3. 

Moment of force, 17, 18. 

Moment of inertia, 69, 71. 

experimental determination of, 191. 

graphical method, 85. 

greatest and least, 76, 89. 

inclined axis, 74, 102. 

non-homogeneous bodies, 102. 

of angle section, 82. 

of circular area, 80. 

of circular cone, 98. 

of elliptical area, 81. 

of locomotive drive wheel, 107. 

of rectangle, 78. 

of triangle, 79. 

parallel axes, 72, 99. 

principal, 104. 

polar, 77. 

right prism, 95. 

Simpson's rule, 88. 

solid of revolution, 97. 

thin plates, 93. 
Motion, curvilinear, 142. 

body through atmosphere, 138. 

due to repulsive force, 134. 

earth, 219. 

harmonic, 132. 

in circle, 146. 

in straight line, 123. 

Newton's laws of, 127. 

on inclined plane, 128. 

resistance varies as distance, 134. 

tAvisted curve, 165. 

Newton's laws of motion, 127. 
Non-concurrent forces, 56, 63. 



Parallel forces, 20, 22. 
Pendulum, compound, 

cycloid al, ir)4. 

simple circular, 148. 



188. 



INDEX 



385 



Percussion, center of, 187, 330. 

Pile driver, 240. 

Pivots, friction of, 299. 

Plane of rotation, 220. 

Polar moment of inertia, 77. 

Power, 233. 

Precessional moment, 222, 225. 

Principal axes, 104. 

Principal moment of inertia, 104. 

ProcUict of inertia, 75. 

Projectile, 156, 160, 162. 

Pulleys and cords, 113. 

Peciprocals of numbers, 359. 
Pectilinear motion, 123. 
Ridative velocity, 139. 
liepresentation of force, 5. 

of couples, 51. 

of moment of inertia, 72. 
Repulsive force, 134. 
Resistance, of roads, 277. 

train, 312. 

varies as distance, 134. 
Ri.2:id body, 2. 

free to rotate, 197. 
Roller beariuij^s, 279. 
Rolling friction, 272. 
Ropes and belts, stiffness of, 294. 
Rotating body, reactions of supports, 

181. 
Rotation, about axis, one point fixed, 
216. 

axis fixed, 247. 

axis not a gravity axis, 201. 

and translation, 173, 208. 

fly wheel, 204. 

in general, 175. 

locomotive drive wheel, 200. 

rigid body, 179. 

sphere, 185. 

symmetrical bodies, 198. 

Side rod of locomotive, 211. 
Simple circular pendulum, 148. 
Simpson's rule, 41, 43. 



Specific gravity, 3. 
Spherical pivot, 302. 
Spinning top, 218. 
Squares, square roots, etc., 359. 
Steam hammer, 244. 
Stiffness of belts, 294. 
Suspension bridge, 114, 120. 

Tangential and normal acceleration. 

144. 
Tangential and normal force, 146. 
Testing of lubricants, 270. 
Theorems of Pai)pus and Guldinus, 47. 
Top, spinning, 218. 
Torsion balance, 192. 
Train resistance, 312. 
Translation and rotation, 173, 208. 
Translation of rigid body, 178. 
Transmissibility of force, 8. 
Transmission dynamometer, 291. 
Trigonometric Functions, 349. 
Twisted curve, motion in, 165. 

Uniform motion in circle, 146. 
Unit of force, 1. 

of moment of inertia, 71. 

of power, 233. 

of weight, 2, 3. 

of work, 230. 

Variable acceleration, 131, 172. 
Varignon's Theorem of Moments, 18. 
Velocity, 123, 142, 169. 
relative, 139. 

Work, combined rotation and transla- 
tion, 254. 

graphical representation, 230. 

motion uniform, 257. 

units of, 230. 

variable force, 238. 
Work and energy, 229. 
AVork-energy relation for any motion, 

257. 
Worn bearing, friction of, 297. 



2c 



Text-Books on Mechanics 



FRANKLIN and MACNUTT — The Elements of Mechanics. A Text-Book 

for Colleges and Technical Schools. By W. S. FRANKLIN and Barry 

Macnutt of Lehigh University. Cloth, Svo, xi + 2Sj pages, $1.50 ?iet. 

Its special aim is to relate the teaching of mechanics to the immediately 

practical things of life, to cultivate suggestiveness w ithout loss of exactitude. 

DUFF — Elementary Experimental Mechanics. By A. Wilmer Dukf, D.Sc. 
(Edin.), Professor of Physics in the Worcester l*olytechnic Institute. New 
York, 1905. Cloih, 26-] pages, $1.60 ?ict. 

LE CONTE — An Elementary Treatise on the Mechanics of Machinery. 
With special reference to the Mechanics of the Steam Engine. By Josi.rii X. 
Le CONTE, Instructor in Mechanical Engineering, University of California; 
Associate Member of the American Institute of Electrical Engineers, etc. 

Cloth, i2mo, $2.2^ net. 

SLATE — The Principles of Mechanics. An Elementary Exposition for Students 
of Physics. By FREDERICK SLATE, Professor of Physics in the University of 
California. Cloth, i2mo, $i.go net. 

The material contained in these chapters has taken on its present form 
gradually, by a process of recasting and sifting. The ideas guiding that pro- 
cess have been three : first, to select the subject-matter with close reference to 
the needs of college students ; second, to bring the instruction into adjustment 
with the actual stage of their training; and, third, to aim continually at treating 
mechanics as a system of organized thought, having a clearly recognizable 
culture value. 

ZIWET — Elements of Theoretical Mechanics. By Alexander Ziwet, Jun- 
ior Professor of Mathematics in the University of Michigan. Revised Edition 
of "An Elementary Treatise on Theoretical Mechanics," especially designed 
for students of engineering. Cloth, Svo, $4.00 net. 

" I can state without hesitation or quahfication that the work is one that is 
unexcelled, and in every way surpasses as a text-book for class use all other 
works on this subject; and, moreover, I find the students all giving it the 
highest praise for the clear and interesting manner in which the subject is 
treated." — M. J. McCUE, M.S., C.E., University of Notre Dame, Ind. 



Carriage on **net'* books Is uniformly an extra charge 



THE MACMILLAN COMPANY 

64-66 FIFTH AVENUE, NEW YORK 
Boston Chicago San Francisco Atlanta 



Standard Books on Mechanics^ etc* 



ABBOT — Problems of the Panama Canal: Including Climatology of the Isth- 
■ mus, Physics and Hydraulics of the River Chagres, Cut at the Continental 
Divide, and a Discussion of the Plans for the Waterway, with History from 
1890 to date. By Hrig.-Gen. HENRY L. Abbot, U.S.A. New Edition. 

C7o^/i, gilt top, 8vo, xil + 2yo pages, index, ^2.00 net. 

BAMFORD — Moving Loads on Railway TJnderbridges. Including Diagrams 
of Bending Moments and Shearing Forces and Tables of Equivalent Uniform 
Live Loads. By HARRY Bamford. Cloth, 8vo, diagrams, $1.2^ net. 

BOYNTON — Application of the Kinetic Theory to Gases, Vapors, Pure 

Liquids, and the Theory of Solutions. By William Pingry Boynton, 
University of Oregon. Cloth, 8vo, 10 + 288 pages, $1.60 ?iet. 

DERR — Photography for Students of Physics and Chemistry. By Louis 
Derr, M.A., S.B., Associate Professor of Physics, Massachusetts Institute of 
Technology. Cloth, crown 8vo, $1.40 net. 

DTJNRAVEN — Self -Instruction in the Practice and Theory of Navigation. 

By the Earl of Dunraven, Extra Master. Enlarged and revised edition. Three 
volumes and supplement. The set, $8.00 net. 

HALLOCK and WADE — Outlines of the Evolution of Weights and Meas- 
ures and the Metric System. By William Hallock, Ph.D., Professor of 
Physics in Columbia University, and Herbert T. Wade. 

Cloth, 8vo, 204 pages, with illustrations^ $2.2^ net. 

HATCH and VALLENTINE — The Weights and Measures of International 
Commerce. Tables and Equivalents. By F. H. Hatch, Ph.D., and F. H. 
Vallentine. Cloth, crown 8vo, 59 pages, $.80 net. 

Mining Tables. Crown 8vo, $i.go net. 

REEVE — The Thermodynamics of Heat-Engines. By Sidney A. Reeve, 
Worcester I^olytechnic Institute. Cloth, i2ino, xi -[■ ji6 pages, $2.60 net. 

SOTHERN — Verbal Notes and Sketches for Marine Engineers. By J. \\\ 
SOTHERN. Fifth Edition, revised and enlarged. 

Cloth, 8vo, XX i + 4JI pages, illustrated, $2.60 net. 

TAYLOR — Resistance of Ships and Screw Propulsion. By D. W. Taylor, 

Naval Constructor, United States Navy. New Edition. 

Cloth, 2J4 pages, diagrams, etc., $2.2^ net. 



Carriage on **uet'* books is uniformly an extra charge 



THE MACMILLAN COMPANY 

64-66 FIFTH AVENUE, NEW YORK 
Boston Chicag-o San Francisco Atlanta 



]^^ 12 1909 



-^ 



